IB DP Math AA Topic: SL 1.5: Laws of exponents with integer; laws of logarithms base 10 and e SL Paper 1

Detailed solution

(a) Find \(\log_{10}(\frac{1}{a})\)

 the given information: \(\log_{10}a = \frac{1}{3}\). This means that \(10^{\frac{1}{3}} = a\), because the logarithm \(\log_{10}a\) is the power to which 10 must be raised to get \(a\). So, \(a = 10^{\frac{1}{3}}\), which is the cube root of 10—neat!

Now, we need \(\log_{10}(\frac{1}{a})\). There’s a handy logarithm property that says \(\log_b\left(\frac{1}{x}\right) = -\log_b x\). Applying this:
\[
\log_{10}\left(\frac{1}{a}\right) = -\log_{10}a
\]
Since \(\log_{10}a = \frac{1}{3}\), we substitute:
\[
\log_{10}\left(\frac{1}{a}\right) = -\frac{1}{3}
\]

Let’s verify this by working it another way. If \(a = 10^{\frac{1}{3}}\), then:
\[
\frac{1}{a} = \frac{1}{10^{\frac{1}{3}}} = 10^{-\frac{1}{3}}
\]
So:
\[
\log_{10}\left(10^{-\frac{1}{3}}\right) = -\frac{1}{3}
\]
This matches perfectly, since \(\log_{10}(10^k) = k\). The answer feels solid!

(b) Find \(\log_{1000}a\)

Now for \(\log_{1000}a\). The base is 1000, which is \(10^3\), so we can use the change of base formula: \(\log_b x = \frac{\log_k x}{\log_k b}\), where we’ll choose base 10 for convenience (since we’re given \(\log_{10}a\)):
\[
\log_{1000}a = \frac{\log_{10}a}{\log_{10}1000}
\]
We know \(\log_{10}1000 = \log_{10}(10^3) = 3\), and \(\log_{10}a = \frac{1}{3}\). Substituting:
\[
\log_{1000}a = \frac{\frac{1}{3}}{3} = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}
\]

Let’s double-check. Since \(a = 10^{\frac{1}{3}}\), we want \(\log_{1000}(10^{\frac{1}{3}})\). If \(\log_{1000}a = x\), then:
\[
1000^x = a
\]
Substitute \(1000 = 10^3\) and \(a = 10^{\frac{1}{3}}\):
\[
(10^3)^x = 10^{\frac{1}{3}}
\]
Using exponent rules, \((10^3)^x = 10^{3x}\), so:
\[
10^{3x} = 10^{\frac{1}{3}}
\]
The exponents must equal:
\[
3x = \frac{1}{3} \quad \Rightarrow \quad x = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}
\]

…………………………Markscheme……………………….

Ans:

(a)  \(log_{10}1-log_{10}a\)  OR  \(log_{10}a^{-1}=-log_{10}a\)  OR  \(log_{10}10^{-\frac{1}{3}}\)  OR  \(10^{x}=\frac{1}{10^{\frac{1}{3}}}\)

=\(-\frac{1}{3}\)

(b) \(\frac{log_{10}a}{log_{10}1000}\)  OR  \(\frac{1}{3}log_{1000}10\)  OR  \(log_{1000}\sqrt[3]{1000^{\frac{1}{3}}}\)  OR \(10^{\frac{1}{3}}=1000^{x}(=(10^{3})^{x})\)

      \(\frac{log_{10}a}{3}\)  OR  \(\frac{1}{3}log_{1000}1000^{\frac{1}{3}}\)  OR \(log_{1000}1000^{\frac{1}{9}}\)  OR  \(3x=\frac{1}{3}\)

=\(\frac{1}{9}\)