Question

Prove that : `4^(log 9) = 3^(log 16)`.

Answer 1

Given: Let “log” denote logarithm to a fixed base the same base everywhere, e.g. base 10.

To Prove: `4^(log 9) = 3^(log 16)`

Proof [Step-wise]:

1. Take log of both sides same log base throughout.

It suffices to show `log(4^(log 9)) = log(3^(log 16))`.

2. Use the power rule for logs:

log(a^b) = b × log a 

So `log(4^(log 9)) = (log 9)(log 4)`

And `log(3^(log 16)) = (log 16)(log 3)`

3. Express 9 and 16 as powers:

9 = 3²

And 16 = 4² 

Therefore log 9 = log (3²)

= 2 × log 3 

And log 16 = log (4²)

= 2 × log 4

4. Substitute into the expressions from step 2:

(log 9)(log 4) = (2 × log 3)(log 4)

= 2 × (log 3)(log 4)

And (log 16)(log 3) = (2 × log 4)(log 3)

= 2 × (log 4)(log 3)

5. The two sides are equal because multiplication is commutative:

2 × (log 3)(log 4)

= 2 × (log 4)(log 3)

6. Hence `log(4^(log 9)) = log(3^(log 16))`. 

Since the logarithm is one-to-one, the original numbers are equal.

`4^(log 9) = 3^(log 16)`