quick exponential question

[paulo1913]

how would I solve this?

 

13^4x-5 = 6

 

because I thought that 13 and 6 would have had to have a common power...

[Cap'n Refsmmat]

Do you mean [math]13^{4x} - 5 = 6[/math] or [math]13^{4x - 5} = 6[/math]?

 

Either way, actually, it's just a logarithm. To get the 4x "down" you need to take the logarithm of both sides.

 

Remember that [math]\log_3 (3^x) = x[/math] and so on.

[paulo1913]

woops i mean 13^(4x-5)=6

 

so would i go log(4x-5)13=log6?

[ydoaPs]

No. You'll also need the change of base formula if your calculator doesn't let you choose your own base.

[paulo1913]

I don't quite understand...

[Cap'n Refsmmat]

You'd have to take the base 13 logarithm of both sides.

 

[math]\log_{13}(13^{4x-5}) = 4x -5[/math]

 

If you calculator can't do base 13 logs (to find [math]\log_{13}(6)[/math]), you can do this:

 

[math]\log_n(a) = \frac{\log_{10}(a)}{\log_{10}(n)}[/math]