exponential equation

[keljane]

This problem is really frustrating me. I don't know what to do after taking the log of each side. I may not be starting in the right place. Can anyone help ?
Find the solution of the exponential equation, correct to four decimal places.
4^5x + 1 = 5^x ? 4

 

[tkhunny]

You'll have to commit on the notation, first.

You have written this: \(\displaystyle 4^{5}\cdot x + 1 = 5^{x} - 4\)

That has two solutions that are not very close together.

Did you mean this? \(\displaystyle 4^{5x+1} = 5^{x-4}\)

 

[keljane]

Yes. That is what I meant. Thank you for correcting me.

 

[tkhunny]

Okay, a logarithm seems like a good idea. Show us how to implement that.

 

[mmm4444bot]

Yes. That is what I [mean].

We text it using grouping symbols around the exponents:

4^(5x + 1) = 5^(x ? 4)