Base-10 logarithms? help

how do I solve 4^(-x+7)=13^(8x) ?

First of all, I'll clear the ground by saying I'll notate log = log(base 10). It is much easier to see.

Onward with the problem!
Make each of the terms on both sides the inside of a log.
Why can we do this?
Just think of it like this: 1 = 1 like log(1) = log(1)

log(4^(-x+7))=log(13^(8x))

Take the exponents out and simplify accordingly
(-x+7)(log(4) = 8xlog(13)
-xlog(4) + 7log(4) = 8xlog(13)

Now solve for x. Put all terms with x on one side and other terms on the other
7log(4) = xlog(4) + 8xlog(13)
7log(4) = x(log(4) + 8log(13))
Divide both sides by log(4) + 8log(13) to find x
x = (7log(4))/(log(4) + 8log(13))

Hope this helped!

Whenever your unknown is in the exponent, you are likely to need to use logs. Take the log of both sides of the equation
log(4^(-x + 7)) = log(13^(8x))
log2^(2(7-x)) = log(13^(8x))
remember your log rules: log(a^b) = b*log(a)
2(7-x) * log(2) = 8x*log(13)
(14-2x) * log(2) = 8x * Log(13)
14log(2) - 2xlog(2) = 8x log(13)
14log(2) = 8x log(13) + 2x log(2)
14log(2) = 2x(4log(13) + log(2))
7log(2) = x(4log(13) + log(2))
log(2^7) = x(log(13^4) + log(2))
log(a) + log(b) = log(ab)
log(2^7) = x *log(13^4 * 2)
x =log(2^7)/ log(13^4 * 2) ≈ 0.442988661

Take the common log of both sides,

(-x + 7)log(4) = 8xlog(13)

-xlog(4) + 7log(4) = 8xlog(13)

x[8log(13) + log(4)] = 7log(4)

x = 7log(4)/[8log(13) + log(4)] = 4.214419939/(8.911546818 + 0.602059991)

= 0.442988661

Check the answer:

4^(-x + 7) = 4^(6.557011339) = 8865.723652

13^(8x) = 13^(3.543909286) = 8865.723652

It checks!

Log 4^(-x+7) = Log 13^(8x)
(-x+7) *Log 4 = (8x) *Log 13
-x+7 = 8x * log 13 / log 4
-x +7 = 14.8 x
7 =1 5.8 X
x = 7 /15.8
X= 35/79