### Question

Find the set of values of x satisfying the inequality 2|2x − a |< |x + 3a |, where a is a positive constant.

To solve the inequality $$2|2x – a| < |x + 3a|$$, where $$a$$ is a positive constant, we’ll need to consider the absolute value expressions separately. Absolute value inequalities can be tricky because they require considering different cases based on the sign of the expressions inside the absolute value.
The inequality $$2|2x – a| < |x + 3a|$$ can be split into different cases depending on the sign of $$2x – a$$ and $$x + 3a$$.
Case 1: $$2x – a \geq 0$$ and $$x + 3a \geq 0$$
When both expressions inside the absolute values are non-negative, we can remove the absolute values. The inequality becomes:
$$2(2x – a) < x + 3a$$
Solving this, we get:
$$4x – 2a < x + 3a$$
$$3x < 5a$$
$$x < \frac{5}{3}a$$
But we also need to consider the original conditions $$2x – a \geq 0$$ and $$x + 3a \geq 0$$. This gives us:
$$2x \geq a \Rightarrow x \geq \frac{a}{2}$$
$$x \geq -3a$$ (which is always true since $$a$$ is positive)
So for Case 1, $$x$$ must satisfy $$\frac{a}{2} \leq x < \frac{5}{3}a$$.
Case 2: $$2x – a < 0$$ and $$x + 3a \geq 0$$
Here, the inequality becomes:
$$-2(2x – a) < x + 3a$$
$$-4x + 2a < x + 3a$$
$$-5x < a$$
$$x > -\frac{a}{5}$$
Again, considering the original conditions, $$2x – a < 0$$ gives us $$x < \frac{a}{2}$$ and $$x + 3a \geq 0$$ is always true.
So for Case 2, $$x$$ must satisfy $$-\frac{a}{5} < x < \frac{a}{2}$$.
Case 3: $$2x – a \geq 0$$ and $$x + 3a < 0$$
This case is not possible because $$a$$ is positive and $$x$$ cannot be both greater than $$\frac{a}{2}$$ and less than $$-3a$$ simultaneously.
Case 4: $$2x – a < 0$$ and $$x + 3a < 0$$
The inequality becomes:
$$-2(2x – a) < -(x + 3a)$$
$$-4x + 2a < -x – 3a$$
$$-3x < -5a$$
$$x > \frac{5}{3}a$$
Considering the original conditions, $$2x – a < 0$$ gives $$x < \frac{a}{2}$$ and $$x + 3a < 0$$ gives $$x < -3a$$. But $$x > \frac{5}{3}a$$ contradicts both of these, so this case does not contribute any solutions.
Combining the Cases
Combining the solutions from Case 1 and Case 2, the set of values of $$x$$ that satisfy the inequality is:
$$-\frac{a}{5} < x < \frac{5}{3}a$$

### Question

Solve the inequality| 2x − 3 |> 4|x + 1|.

State or imply non-modular inequality$$(2x-3)^{2}>4^{2}(x+1)$$

To solve the inequality $$|2x – 3| > 4|x + 1|$$, we can indeed square both sides of the inequality to eliminate the absolute values. This step is valid because both sides of the inequality are non-negative, and squaring is a monotonic function in the non-negative domain.
Starting with the original inequality:
$$|2x – 3| > 4|x + 1|$$
Square both sides:
$$(|2x – 3|)^2 > (4|x + 1|)^2$$
$$(2x – 3)^2 > 16(x + 1)^2$$
Expanding both sides:
$$(2x – 3)^2 > 16(x^2 + 2x + 1)$$
$$4x^2 – 12x + 9 > 16x^2 + 32x + 16$$
Bringing all terms to one side of the inequality:
$$4x^2 – 16x^2 – 12x – 32x + 9 – 16 < 0$$
$$-12x^2 – 44x – 7 < 0$$
To solve this quadratic inequality, we need to find the roots of the corresponding quadratic equation $$-12x^2 – 44x – 7 = 0$$. The roots can be found using the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
where $$a = -12$$, $$b = -44$$, and $$c = -7$$.
$$x = \frac{-(-44) \pm \sqrt{(-44)^2 – 4(-12)(-7)}}{2(-12)}$$
$$x = \frac{44 \pm \sqrt{1936 – 336}}{-24}$$
$$x = \frac{44 \pm \sqrt{1600}}{-24}$$
$$x = \frac{44 \pm 40}{-24}$$
This gives two roots:
$$x_1 = \frac{44 + 40}{-24}$$
$$x_2 = \frac{44 – 40}{-24}$$
$$x_1=-3.5$$
$$x_2=-0.16$$
$$\therefore -3.5< x< -0.16$$

### Question

Solve the equation $$2\left | 3^{x}-1 \right |=3^{x}$$ , giving your answers correct to 3 significant figures.

To solve the equation $$2|3^x – 1| = 3^x$$, let’s first deal with the absolute value. We have two cases to consider:
Case 1: $$3^x – 1 \geq 0$$
In this case, $$|3^x – 1| = 3^x – 1$$, and the equation becomes:
$$2(3^x – 1) = 3^x$$
$$2 \cdot 3^x – 2 = 3^x$$
$$2 \cdot 3^x – 3^x = 2$$
$$3^x = 2$$
Taking the logarithm of both sides (base 3):
$$x = \log_3(2)$$
Case 2: $$3^x – 1 < 0$$
In this case, $$|3^x – 1| = 1 – 3^x$$, and the equation becomes:
$$2(1 – 3^x) = 3^x$$
$$2 – 2 \cdot 3^x = 3^x$$
$$2 = 5 \cdot 3^x$$
Dividing both sides by 5:
$$\frac{2}{5} = 3^x$$
Taking the logarithm of both sides (base 3):
$$x = \log_3\left(\frac{2}{5}\right)$$
Now, let’s calculate these logarithmic values to three significant figures.
For $$x = \log_3(2)$$:
$$x \approx \log_3(2) \approx \frac{\ln(2)}{\ln(3)}$$
For $$x = \log_3\left(\frac{2}{5}\right)$$:
$$x \approx \log_3\left(\frac{2}{5}\right) \approx \frac{\ln(2/5)}{\ln(3)}$$
x=0.631 and x =-0.834

### Question

(i) Solve the equation $$\left | 4x-1 \right |=\left | x-3 \right |.$$

(ii) Hence solve the equation $$\left | 4^{y+1}-1 \right |=\left | 4^{y}-3 \right |$$ correct to 3 significant figures.

(i): Solve $$|4x – 1| = |x – 3|$$
To solve this equation, we consider different cases based on the values within the absolute value brackets.
Case 1: $$4x – 1 \geq 0$$ and $$x – 3 \geq 0$$
When both expressions are non-negative, the equation becomes:
$$4x – 1 = x – 3$$
Solving for $$x$$:
$$4x – x = -3 + 1$$
$$3x = -2$$
$$x = -\frac{2}{3}$$
However, this solution does not satisfy the original conditions of this case (both $$4x – 1$$ and $$x – 3$$ being non-negative), so we discard it.
Case 2: $$4x – 1 < 0$$ and $$x – 3 < 0$$
When both expressions are negative, the equation becomes:
$$-(4x – 1) = -(x – 3)$$
$$-4x + 1 = -x + 3$$
$$-4x + x = 3 – 1$$
$$-3x = 2$$
$$x = -\frac{2}{3}$$
Again, this solution does not satisfy the conditions of this case (both $$4x – 1$$ and $$x – 3$$ being negative), so we discard it.
Case 3: $$4x – 1 \geq 0$$ and $$x – 3 < 0$$
This case has mixed signs, so the equation becomes:
$$4x – 1 = -(x – 3)$$
$$4x – 1 = -x + 3$$
$$4x + x = 3 + 1$$
$$5x = 4$$
$$x = \frac{4}{5}$$
This solution $$x = \frac{4}{5}$$ satisfies the conditions of this case.
Case 4: $$4x – 1 < 0$$ and $$x – 3 \geq 0$$
The equation becomes:
$$-(4x – 1) = x – 3$$
$$-4x + 1 = x – 3$$
$$-4x – x = -3 – 1$$
$$-5x = -4$$
$$x = \frac{4}{5}$$
This solution $$x = \frac{4}{5}$$ also satisfies the conditions of this case.
So, the solution to $$|4x – 1| = |x – 3|$$ is $$x = \frac{4}{5}$$.
(ii): Solve $$\left|4^{y+1}-1\right|=\left|4^y-3\right|$$
Given the solution to the first part, we can rewrite the second equation in a similar form:
$$\left|4^{y+1}-1\right|=\left|4^y-3\right|$$
$$|4 \cdot 4^y – 1| = |4^y – 3|$$
Using the solution from part (i), where $$|4x – 1| = |x – 3|$$ had the solution $$x = \frac{4}{5}$$, we can set $$4^y = x$$ and solve for $$y$$
$$4^y = \frac{4}{5}$$
Taking the logarithm base 4 of both sides:
$$y = \log_4\left(\frac{4}{5}\right)$$
$$y=-0.1609$$

### Question

Solve the inequality |2x + 1| <| 3x − 2|.

To solve the inequality $$|2x + 1| < |3x – 2|$$, we need to consider the different cases based on the values inside the absolute values. The absolute value expressions represent the distance from the origin on a number line, so we are comparing these distances in each case.
Case 1: $$2x + 1 \geq 0$$ and $$3x – 2 \geq 0$$
When both expressions inside the absolute values are non-negative, the inequality becomes:
$$2x + 1 < 3x – 2$$
Solving this inequality:
$$2x – 3x < -2 – 1$$
$$-x < -3$$
$$x > 3$$
But we also need to consider the conditions of this case. $$2x + 1 \geq 0$$ implies $$x \geq -\frac{1}{2}$$ and $$3x – 2 \geq 0$$ implies $$x \geq \frac{2}{3}$$. So for Case 1, $$x$$ must satisfy $$x > 3$$.
Case 2: $$2x + 1 < 0$$ and $$3x – 2 < 0$$
When both expressions are negative, the inequality becomes:
$$-(2x + 1) < -(3x – 2)$$
$$-2x – 1 < -3x + 2$$
$$-2x + 3x < 2 + 1$$
$$x < 3$$
Considering the original conditions, $$2x + 1 < 0$$ implies $$x < -\frac{1}{2}$$ and $$3x – 2 < 0$$ implies $$x < \frac{2}{3}$$. So for Case 2, $$x$$ must satisfy $$x < -\frac{1}{2}$$.
Case 3: $$2x + 1 \geq 0$$ and $$3x – 2 < 0$$
This case has mixed signs, so the inequality becomes:
$$2x + 1 < -(3x – 2)$$
$$2x + 1 < -3x + 2$$
$$2x + 3x < 2 – 1$$
$$5x < 1$$
$$x < \frac{1}{5}$$
Considering the conditions, $$2x + 1 \geq 0$$ gives $$x \geq -\frac{1}{2}$$ and $$3x – 2 < 0$$ gives $$x < \frac{2}{3}$$. So for Case 3, $$x$$ must satisfy $$-\frac{1}{2} \leq x < \frac{1}{5}$$.
Case 4: $$2x + 1 < 0$$ and $$3x – 2 \geq 0$$
The inequality becomes:
$$-(2x + 1) < 3x – 2$$
$$-2x – 1 < 3x – 2$$
$$-2x – 3x < -2 + 1$$
$$-5x < -1$$
$$x > \frac{1}{5}$$
Considering the conditions, $$2x + 1 < 0$$ gives $$x < -\frac{1}{2}$$ and $$3x – 2 \geq 0$$ gives $$x \geq \frac{2}{3}$$. But $$x > \frac{1}{5}$$ contradicts $$x < -\frac{1}{2}$$, so this case does not contribute any solutions.
Combining the Cases
Combining the solutions from Cases 1, 2, and 3, the set of values of $$x$$ that satisfy the inequality is $$x < -\frac{1}{2}$$ and $$-\frac{1}{2} \leq x < \frac{1}{5}$$ and $$x > 3$$.
So, the final solution is:
$$x < -\frac{1}{2} \quad \text{or} \quad -\frac{1}{2} \leq x < \frac{1}{5} \quad \text{or} \quad x > 3$$

### Question

Solve the inequality 2 − 5x > 2|x − 3|.

To solve the inequality $$2 – 5x > 2|x – 3|$$, we need to consider the different cases based on the value inside the absolute value. Let’s handle it step by step.
Case 1: $$x – 3 \geq 0$$
When $$x – 3$$ is non-negative, the inequality becomes:
$$2 – 5x > 2(x – 3)$$
$$2 – 5x > 2x – 6$$
$$-5x – 2x > -6 – 2$$
$$-7x > -8$$
$$x < \frac{8}{7}$$
But we also need to consider the original condition $$x – 3 \geq 0$$, which gives us $$x \geq 3$$. Since $$x < \frac{8}{7}$$ and $$x \geq 3$$ are contradictory, there are no solutions in this case.
Case 2: $$x – 3 < 0$$
When $$x – 3$$ is negative, the absolute value expression becomes $$3 – x$$, and the inequality becomes:
$$2 – 5x > 2(3 – x)$$
$$2 – 5x > 6 – 2x$$
$$-5x + 2x > 6 – 2$$
$$-3x > 4$$
$$x < -\frac{4}{3}$$
Considering the original condition $$x – 3 < 0$$, which gives us $$x < 3$$, this solution $$x < -\frac{4}{3}$$ is valid within this range.
Combining the Cases
The only valid solution comes from Case 2, so the set of values of $$x$$ that satisfy the inequality $$2 – 5x > 2|x – 3|$$ is:
$$x < -\frac{4}{3}$$

### Question

(a) Sketch the graph of y = |x-2|
(b) Solve the inequality |x-2|<3x-4.

(a) Sketch the graph of $$y = |x – 2|$$
The graph of $$y = |x – 2|$$ is a V-shaped graph with its vertex at $$x = 2$$, where the graph changes direction.
1. For $$x \geq 2$$, $$|x – 2| = x – 2$$. The graph in this region is a straight line with a slope of 1, starting from the point (2,0) and going upwards to the right.
2. For $$x < 2$$, $$|x – 2| = 2 – x$$. The graph in this region is a straight line with a slope of -1, starting from the point (2,0) and going upwards to the left.
The graph intersects the y-axis at $$y = 2$$ (when $$x = 0$$).
Here is the graph of $$y = |x – 2|$$. As described, it’s a V-shaped graph with its vertex at $$x = 2$$, intersecting the y-axis at $$y = 2$$. The graph shows the two distinct linear parts of the absolute value function, one increasing and the other decreasing.

(b) Solve the inequality $$|x – 2| < 3x – 4$$
To solve this inequality, let’s consider the two cases of the absolute value:
Case 1: $$x – 2 \geq 0$$ or $$x \geq 2$$
When $$x \geq 2$$, $$|x – 2| = x – 2$$. The inequality becomes:
$$x – 2 < 3x – 4$$
$$-2x < -2$$
$$x > 1$$
Since we are considering $$x \geq 2$$, the solution for this case is $$x > 2$$.
Case 2: $$x – 2 < 0$$ or $$x < 2$$
When $$x < 2$$, $$|x – 2| = 2 – x$$. The inequality becomes:
$$2 – x < 3x – 4$$
$$-4x < -6$$
$$x > \frac{3}{2}$$
Considering $$x < 2$$, the solution for this case is $$\frac{3}{2} < x < 2$$.
Combining the Cases
Combining the solutions from both cases, the solution to the inequality $$|x – 2| < 3x – 4$$ is $$x > \frac{3}{2}$$.
Note that $$x > 2$$ and $$\frac{3}{2} < x < 2$$ overlap, so the final solution is simply $$x > \frac{3}{2}$$.

### Question

Solve the equation $$(1+2^{x})=2$$, giving your answer correct to 3 decimal places.

Remove logarithm and obtain $$1+2^{x}=e^{2}$$

Use correct method to solve an equation of the form $$2^{x}$$ = a , where a > 0Obtain answer x = 2.676

### Question

Solve the inequality| x − 4| < 2|3x + 1|.

To address the inequality $$|x – 4| < 2|3x + 1|$$ as per your instructions, we can consider the non-modular inequality $$(x – 4)^2 < (2(3x + 1))^2$$ and the corresponding pair of linear equations $$x – 4 = \pm 2(3x + 1)$$.
Non-Modular Inequality: $$(x – 4)^2 < (2(3x + 1))^2$$
$$(x – 4)^2 < (6x + 2)^2$$
$$x^2 – 8x + 16 < 36x^2 + 24x + 4$$
$$0 < 36x^2 + 24x + 4 – x^2 + 8x – 16$$
$$0 < 35x^2 + 32x – 12$$
3. Solve for $$x$$ using the quadratic formula or factorization.
Pair of Linear Equations: $$x – 4 = \pm 2(3x + 1)$$
Equation 1: $$x – 4 = 2(3x + 1)$$
Expand and solve for $$x$$:
$$x – 4 = 6x + 2$$
$$-5x = 6$$
$$x = -\frac{6}{5}$$
Equation 2: $$x – 4 = -2(3x + 1)$$
Expand and solve for $$x$$:
$$x – 4 = -6x – 2$$
$$7x = 2$$
$$x = \frac{2}{7}$$
Combining the Solutions
Also,
Solve $$35x^2 + 32x – 12 = 0$$. We can use the quadratic formula, $$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$, where $$a = 35$$, $$b = 32$$, and $$c = -12$$.
The roots are:
$$x_{1,2} = \frac{-32 \pm \sqrt{32^2 – 4 \cdot 35 \cdot (-12)}}{2 \cdot 35}$$
Calculate the discriminant and then the roots:
$$x_{1,2} = \frac{-32 \pm \sqrt{1024 + 1680}}{70}$$
$$x_{1,2} = \frac{-32 \pm \sqrt{2704}}{70}$$
$$x_{1,2} = \frac{-32 \pm 52}{70}$$
$$x_1 = \frac{-32 + 52}{70} = \frac{20}{70} = \frac{2}{7}$$
$$x_2 = \frac{-32 – 52}{70} = \frac{-84}{70} = -\frac{6}{5}$$
The roots divide the number line into three intervals: $$x < -\frac{6}{5}$$, $$-\frac{6}{5} < x < \frac{2}{7}$$, and $$x > \frac{2}{7}$$. Test points from each interval in the original inequality to determine where it holds true.
By testing these intervals, we find that the inequality $$0 < 35x^2 + 32x – 12$$ is true for $$x < -\frac{6}{5}$$ and $$x > \frac{2}{7}$$.
Thus, the solution to the inequality is $$x < -\frac{6}{5}$$ or $$x > \frac{2}{7}$$.
The critical values obtained from solving the linear equations are $$x = -\frac{6}{5}$$ and $$x = \frac{2}{7}$$. When considering the quadratic inequality $$0 < 35x^2 + 32x – 12$$, we should test the intervals defined by these critical values to see where the inequality holds true.
After testing these intervals (less than $$-\frac{6}{5}$$, between $$-\frac{6}{5}$$ and $$\frac{2}{7}$$, and greater than $$\frac{2}{7}$$), we find that the solution to the inequality $$|x – 4| < 2|3x + 1|$$ is:
$$x < -\frac{6}{5}$$ and $$x > \frac{2}{7}$$

### Question

Solve the equation 4|5x − 1| = 5x, giving your answers correct to 3 decimal places.

To solve the equation $$4|5^x – 1| = 5^x$$, we will consider two cases based on the absolute value expression.
Case 1: $$5^x – 1 \geq 0$$
This corresponds to $$5^x \geq 1$$. For this case, the equation simplifies to:
$$4(5^x – 1) = 5^x$$
$$4 \cdot 5^x – 4 = 5^x$$
$$4 \cdot 5^x – 5^x = 4$$
$$3 \cdot 5^x = 4$$
$$5^x = \frac{4}{3}$$
$$\log(5^x) = \log\left(\frac{4}{3}\right)$$
Use the property of logarithms ($$\log(b^c) = c \log(b)$$):
$$x \log(5) = \log\left(\frac{4}{3}\right)$$
$$x = \frac{\log\left(\frac{4}{3}\right)}{\log(5)}=0.1787$$
Case 2: $$5^x – 1 < 0$$
This corresponds to $$5^x < 1$$. For this case, the equation simplifies to:
$$4(1 – 5^x) = 5^x$$
$$4 – 4 \cdot 5^x = 5^x$$
$$4 = 5^x + 4 \cdot 5^x$$
$$4 = 5 \cdot 5^x$$
$$5^x = \frac{4}{5}$$
$$\log(5^x) = \log\left(\frac{4}{5}\right)$$
$$x \log(5) = \log\left(\frac{4}{5}\right)$$
$$x = \frac{\log\left(\frac{4}{5}\right)}{\log(5)}=-0.1387$$
In both cases, we obtain an expression for $$x$$ which can be solved using a calculator to find the values correct to three decimal places. Since I can’t perform the calculation here, you can use a scientific calculator to find the values of $$x$$ for both cases. Remember, the valid solution must satisfy the original condition of each case (whether $$5^x \geq 1$$ or $$5^x < 1$$).

### Question

Solve the inequality 2|3x − 1|< |x + 1|.

To solve the inequality $$2|3x – 1| < |x + 1|$$, we need to consider different cases based on the absolute value expressions.
Case 1: $$3x – 1 \geq 0$$ and $$x + 1 \geq 0$$
For this case, $$3x – 1 \geq 0$$ implies $$x \geq \frac{1}{3}$$, and $$x + 1 \geq 0$$ implies $$x \geq -1$$. The inequality becomes:
$$2(3x – 1) < x + 1$$
$$6x – 2 < x + 1$$
$$5x < 3$$
$$x < \frac{3}{5}$$
Combine this with the initial conditions $$x \geq \frac{1}{3}$$ and $$x \geq -1$$. The solution for Case 1 is $$\frac{1}{3} \leq x < \frac{3}{5}$$.
Case 2: $$3x – 1 \geq 0$$ and $$x + 1 < 0$$
For this case, $$3x – 1 \geq 0$$ implies $$x \geq \frac{1}{3}$$, and $$x + 1 < 0$$ implies $$x < -1$$. The inequality becomes:
$$2(3x – 1) < -(x + 1)$$
$$6x – 2 < -x – 1$$
$$7x < 1$$
$$x < \frac{1}{7}$$
However, this does not satisfy the initial conditions $$x \geq \frac{1}{3}$$ and $$x < -1$$, so there is no solution in this case.
Case 3: $$3x – 1 < 0$$ and $$x + 1 \geq 0$$
For this case, $$3x – 1 < 0$$ implies $$x < \frac{1}{3}$$, and $$x + 1 \geq 0$$ implies $$x \geq -1$$. The inequality becomes:
$$2(-(3x – 1)) < x + 1$$
$$-6x + 2 < x + 1$$
$$-7x < -1$$
$$x > \frac{1}{7}$$
Combine this with the initial conditions $$x < \frac{1}{3}$$ and $$x \geq -1$$. The solution for Case 3 is $$\frac{1}{7} < x < \frac{1}{3}$$.
Case 4: $$3x – 1 < 0$$ and $$x + 1 < 0$$
For this case, $$3x – 1 < 0$$ implies $$x < \frac{1}{3}$$, and $$x + 1 < 0$$ implies $$x < -1$$. The inequality becomes:
$$2(-(3x – 1)) < -(x + 1)$$
$$-6x + 2 < -x – 1$$
$$-5x < -3$$
$$x > \frac{3}{5}$$
However, this does not satisfy the initial conditions $$x < \frac{1}{3}$$ and $$x < -1$$, so there is no solution in this case.
Combining All Cases
The solutions from all the cases must be combined to find the overall solution:
– From Case 1: $$\frac{1}{3} \leq x < \frac{3}{5}$$
– From Case 3: $$\frac{1}{7} < x < \frac{1}{3}$$
Therefore, the solution to the inequality $$2|3x – 1| < |x + 1|$$ is $$\frac{1}{7} < x < \frac{3}{5}$$.

### Question

(i) Solve the equation |2x − 1| = 3x.

(ii) Hence solve the equation $$|2|5^{x}-1|=3|5^{x}|$$,giving your answer correct to 3 significant figures.

(i) Solve $$|2x – 1| = 3x$$
To solve this, we consider the non-modular equation $$(2x – 1)^2 = (3x)^2$$ and the pair of linear equations $$2x – 1 = \pm 3x$$.
Non-Modular Equation:
$$(2x – 1)^2 = 9x^2$$
$$4x^2 – 4x + 1 = 9x^2$$
$$0 = 5x^2 + 4x – 1$$
$$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
$$x = \frac{-4 \pm \sqrt{16 + 20}}{10}$$
$$x = \frac{-4 \pm \sqrt{36}}{10}$$
$$x = \frac{-4 \pm 6}{10}$$
The two solutions are:
$$x = -1\ \text{and}\ x = \frac{2}{5}$$
However, since $$-1$$ does not satisfy the original absolute value equation, the only valid solution from this method is $$x = \frac{2}{5}$$.
(ii) Solve $$|2|5^x – 1| = 3|5^x|$$
Given the solutions from part (i), we substitute $$5^x$$ in place of $$x$$:
For $$x = \frac{1}{5}$$:
$$5^x = \frac{1}{5}$$
$$x = -1$$ (as found in part (i))
For $$x = \frac{2}{5}$$:
$$5^x = \frac{2}{5}$$
Now we solve for $$x$$:
$$5^x = \frac{2}{5}$$
$$x = \log_5\left(\frac{2}{5}\right)$$
$$x = \frac{\log\left(\frac{2}{5}\right)}{\log(5)}$$
$$x \approx -0.569$$.

### Question

Solve the equation $$|4-2^{x}|$$ =10, giving your answer correct to 3 significant figures.

To solve the equation $$|4 – 2^x| = 10$$, we consider two cases based on the absolute value.
Case 1: $$4 – 2^x \geq 0$$
For this case, the equation becomes:
$$4 – 2^x = 10$$
Rearranging and solving for $$x$$:
$$2^x = 4 – 10$$
$$2^x = -6$$
However, $$2^x$$ is always positive, so this case yields no solution.
Case 2: $$4 – 2^x < 0$$
For this case, we consider the expression inside the absolute value as negative, and the equation becomes:
$$2^x – 4 = 10$$
$$2^x = 14$$
To find $$x$$, take the logarithm (base 2) of both sides:
$$x = \log_2(14)$$
$$x=3.81$$

### Question

Find the set of values of x satisfying the inequality 3|x − 1| < |2x + 1|.

To solve the inequality $$3|x – 1| < |2x + 1|$$, we need to consider different cases based on the absolute value expressions.
Case 1: $$x – 1 \geq 0$$ and $$2x + 1 \geq 0$$
For this case, $$x – 1 \geq 0$$ implies $$x \geq 1$$, and $$2x + 1 \geq 0$$ implies $$x \geq -\frac{1}{2}$$. The inequality becomes:
$$3(x – 1) < 2x + 1$$
$$3x – 3 < 2x + 1$$
$$x < 4$$
Combine this with the initial conditions $$x \geq 1$$ and $$x \geq -\frac{1}{2}$$.
The solution for Case 1 is $$1 \leq x < 4$$.
Case 2: $$x – 1 \geq 0$$ and $$2x + 1 < 0$$
For this case, $$x – 1 \geq 0$$ implies $$x \geq 1$$, and $$2x + 1 < 0$$ implies $$x < -\frac{1}{2}$$. The inequality becomes:
$$3(x – 1) < -(2x + 1)$$
$$3x – 3 < -2x – 1$$
$$5x < 2$$
$$x < \frac{2}{5}$$
However, this does not satisfy the initial conditions $$x \geq 1$$ and $$x < -\frac{1}{2}$$, so there is no solution in this case.
Case 3: $$x – 1 < 0$$ and $$2x + 1 \geq 0$$
For this case, $$x – 1 < 0$$ implies $$x < 1$$, and $$2x + 1 \geq 0$$ implies $$x \geq -\frac{1}{2}$$. The inequality becomes:
$$3(1 – x) < 2x + 1$$
$$3 – 3x < 2x + 1$$
$$5x > 2$$
$$x > \frac{2}{5}$$
Combine this with the initial conditions $$x < 1$$ and $$x \geq -\frac{1}{2}$$. The solution for Case 3 is $$\frac{2}{5} < x < 1$$.
Case 4: $$x – 1 < 0$$ and $$2x + 1 < 0$$
For this case, $$x – 1 < 0$$ implies $$x < 1$$, and $$2x + 1 < 0$$ implies $$x < -\frac{1}{2}$$. The inequality becomes:
$$3(1 – x) < -(2x + 1)$$
$$3 – 3x < -2x – 1$$
$$-x < -4$$
$$x > 4$$
However, this does not satisfy the initial conditions $$x < 1$$ and $$x < -\frac{1}{2}$$, so there is no solution in this case.
Combining All Cases
The solutions from all the cases must be combined to find the overall solution:
– From Case 1: $$1 \leq x < 4$$
– From Case 3: $$\frac{2}{5} < x < 1$$
Therefore, the solution to the inequality $$3|x – 1| < |2x + 1|$$ is $$\frac{2}{5} < x < 4$$.

### Question

Solve the inequality |2x + 3| > 3|x + 2|.

To address the inequality $$|2x + 3| > 3|x + 2|$$ as per your instructions, we consider the non-modular inequality $$(2x + 3)^2 > 3^2(x + 2)^2$$ and the corresponding pair of linear equations.
Non-Modular Inequality: $$(2x + 3)^2 > 9(x + 2)^2$$
Expand both sides:
$$(2x + 3)^2 > 9(x + 2)^2$$
$$4x^2 + 12x + 9 > 9x^2 + 36x + 36$$
$$0 > 5x^2 + 24x + 27$$
Solve for $$x$$ using the quadratic formula or factorization:
The quadratic formula is $$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$, where $$a = 5$$, $$b = 24$$, and $$c = 27$$.
$$x_{1,2} = \frac{-24 \pm \sqrt{24^2 – 4 \cdot 5 \cdot 27}}{2 \cdot 5}$$
Corresponding Pair of Linear Equations:
1. $$2x + 3 = 3(x + 2)$$ and $$2x + 3 = -3(x + 2)$$
For $$2x + 3 = 3(x + 2)$$:
$$2x + 3 = 3x + 6$$
$$x = -3$$
For $$2x + 3 = -3(x + 2)$$:
$$2x + 3 = -3x – 6$$
$$5x = -9$$
$$x = -\frac{9}{5}$$
Combining Solutions
The critical values obtained from solving the linear equations are $$x = -3$$ and $$x = -\frac{9}{5}$$. When considering the quadratic inequality $$0 > 5x^2 + 24x + 27$$, we should test the intervals defined by these critical values to see where the inequality holds true.
Solution Intervals
1. For $$x < -3$$, the inequality does not hold as it contradicts the quadratic inequality.
2. For $$-3 < x < -\frac{9}{5}$$, the inequality holds.
3. For $$x > -\frac{9}{5}$$, the inequality does not hold as it contradicts the quadratic inequality.
Thus, the solution to the inequality $$|2x + 3| > 3|x + 2|$$ is $$-3 < x < -\frac{9}{5}$$.

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