Proving the Equality of Logarithms with Different Bases

[TN17]

Homework Statement

How would I show that 1/log_ab = log_ba ?

The Attempt at a Solution

I'm not really sure where to start because of the different bases.

 

[Tide]

Hint: What is b^{\log_b a}?

 

[TN17]

Hint: What is b^{\log_b a}?

Is that Power Laws of Logs or Techniques for solving exponential equations, because I haven't learned those yet.
We haven't learned anything with logs as exponents yet.

 

[Tide]

That would be a fundamental identity for logarithms and should have been the first thing you learned about them. Basically, exponentials and logarithms are inverse functions of each other. Check with your textbook. :)

 

[TN17]

That would be a fundamental identity for logarithms and should have been the first thing you learned about them. Basically, exponentials and logarithms are inverse functions of each other. Check with your textbook. :)

Yes, we learned how to write an exponentials as a log and vice versa, and we learned how to evaulate logs, like log₂8 where x=3. But we haven't seen anything with "log___" as the actual exponent before. :S

 

[Tide]

What you need to know is that

\log_b b^a = a and b^{\log_b a} = a

 

[TN17]

What you need to know is that

\log_b b^a = a and b^{\log_b a} = a

I found this online:
"Remember that log_b a = log a / log b for any base/subscript.

Then 1 / [log_a b] = 1 / [log b / log a]

Now dividing by a fraction is the same as mulitplying its reciproical.

1 * (log a / log b)

log a / log b

Now use the fact at the top to change this into
log a / log b = log_b a
Therefore, 1 / [log_a b] = log_b a."

Is that the same thing you were talking about?

We have a quiz on:
Exponential function + its inverse
Logarithms (simple problems)
Transformations of log functions
Using log scales in physical sciences (ex: pH scale)
and Exponential equations.

I'm familiar with all of those, but when I overheard someone from the other class talking about this question right after their quiz, I was curious, so that's why I asked...

 

[vela]

I think Tide was going for this approach:

Start with b^{\log_b a} = a and then take the log, base a, of both sides to get

\log_a (b^{\log_b a}) = \log_a a

Use a property of logs to bring the exponent down, do a little algebra, and you'll get the result you wanted.

The way you did it is fine as well. There are often multiple ways to show the same thing.

 

[Tide]

Vela,

Exactly! I had assumed the original poster hadn't seen the base conversion yet.