Question

How do I solve: 2^(4x) = 9^(x-1)?

Answer 1

ln

Answer 2

\[4x \ln 2=(x-1)\ln 9\]
\[\frac{ \ln 2 }{ \ln 9 }=\frac{ x-1 }{ 4x }\]

Answer 3

this is what I did so far:

4x(x-1) = log9/log2
4x^2-4x = log9/log2
4x^2-4x-(log9/log2)=0?

Answer 4

and then what happens? o.O

Answer 5

\[\frac{ln2}{ln9}4x-x=-1\]
\[x(\frac{ln2}{ln9}4-1)=-1\]
\[x=\frac{-1}{(\frac{ln2}{ln9}4-1)}\]

Answer 6

basically throw all the xs to one side, factor the x, and then divide the other side by what did not get factored by x.

Answer 7

yes

Answer 8

ohh, i see. alright, thank you!

Answer 9

ALTERNATIVELY\[(2^4)^x=9^x9^{-1}\]
\[(\frac{ 2^4 }{ 9 })^x=\frac{ 1 }{ 9 }\]
\[x \rightarrow \log_{\frac{ 16 }{ 9 }}\frac{ 1 }{ 9 }\]