Review question

# Can we find bounds for $\log_2 3$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9381

## Solution

Observe that $2^3 = 8$, $2^5 = 32$, $3^2 = 9$ and $3^3 = 27$. From these facts, we can deduce that $\log_{2}3$, the logarithm of $3$ to base $2$, is

1. between $1\frac{1}{3}$ and $1\frac{1}{2}$;

2. between $1\frac{1}{2}$ and $1\frac{2}{3}$;

3. between $1\frac{2}{3}$ and $2$;

4. between $2$ and $3$.

We are told that $2^3 < 3^2 < 3^3 < 2^5.$

As $\log_{2}x$ is an increasing function (which means that $\log_2x < \log_2y$ if and only if $x < y$, or the gradient of the graph $y = \log_2x$ is always positive), we can take logarithms to base $2$ of these inequalities. This gives $3 < 2\log_{2}3 < 3\log_{2}3 < 5.$

From the first inequality we get $\frac{3}{2} < \log_{2}3,$ and from the last inequality we get $\log_{2}3 < \frac{5}{3}.$

Combining these gives $\frac{3}{2} < \log_{2}3 < \frac{5}{3},$ so the answer is (b).

An alternative approach might say, let $\log_23 = k$, so $3 = 2^k$. Now $9 = 2^{2k} > 8$, so $2k > 3$.

We also have $27 = 2^{3k} < 32 = 2^5$, so $3k < 5$.

We use here that $y = 2^x$ is an increasing function.