How do you prove log(a^x) = 1/log(x^a)

Question

My teacher will give me extra credit if I can prove that

 

log(a^x) = 1/log(x^a)

 

Can you help? Thank you.

Answer 1

You can use the change of base formula to prove it quickly.

log_ax = lnx/lna

1/log_xa = 1/(lna/lnx) = lnx/lna

 

Therefore, log_ax = 1/log_xa.

Answer 2

Use the definition of logarithm:

 

y = log_ax ⇔ a^y=x   (a>0, x>0)

 

Now solve for a:

 

a = x^1/y

 

Now use the definition of logarithm again:

 

a = x^1/y ⇔ 1/y = log_xa, or  y= 1/log_xa

 

Therefore,

 

log_ax = 1/log_xa.

Answer 3

This is not a true statement. Use the same values for x and a to see why.

 

I think what you might have meant is log_ax = 1/log_xa, (where the a on the left-hand side and the x on the right-hand side are the bases of the logarithms, not the bases of an exponent) which is true and can be proven. There is a property of logarithms which states that we can change the base of a logarithm to 10 or e, which are much easier to work with.

 

For any logarithm with any base, log_yz = (log z)/(log y) = (ln z)/(ln y)

 

We can show this using simple numbers. Let's say y is 100 and z is 10. log₁₀₀10 looks daunting, but we can change it to base 10 using this change-of-base formula:

 

log₁₀₀10 = (log 10)/(log 100) = 1/2

 

When we rewrite the original log here, we are asked to find what exponent 100 is raised to in order to get 10. In fact, 100^1/2 = 10, so the change-of-base formula works.

 

Now we can use this to prove the statement above, which is essentially what this change-of-base implies:

 

 

log_ax = (log x)/(log a)          or          log_ax = (ln x)/(ln a)

log_xa = (log a)/(log x)          or          log_xa = (ln a)/(ln x)

 

Regardless of which base we change it to, it is clear that these are inverses of each other, that is, log_ax = 1/log_xa.