5^{-8x}=16^{x+8}
Write the exact answer using base-10 logarithms.

Question

5^{-8x}=16^{x+8} 
Write the exact answer using base-10 logarithms.

anonymous · Answer

$$(-8x)(\log5) = (x+8)(\log16)$$

anonymous · Answer

$$\frac{ -8x }{ x+8 }=\frac{ \log16 }{ \log 5 }$$

anonymous · Answer

So I'm Stuck there

tkhunny · Answer

Yes, you are.  Don't do that.

$(-8x)(log(5)) = (x+8)(log(16)) = xlog(16) + 8log(16)$

Then $(-8x)log(5) - xlog(16) = 8log(16)$

Now what?

anonymous · Answer

Still stuck. Need to solve for x. I cant see what I could move over from there

tkhunny · Answer

Have you considered the distributive property?

$x[(-8)log(5) - log(16)] = 8log(16)$

anonymous · Answer

I had not. Thank you.

tkhunny · Answer

Actually, you could have gotten there the way your had it.

$\dfrac{-8x}{x+8} = -8 - \dfrac{64}{x+8} = \dfrac{log(16)}{log(5)}$

It's no less annoying that the other way we did it already.  It's just another way.

anonymous · Answer

Thanks mate

tkhunny · Answer

Whoops!  "+64", not "-64"!