Question

Find the exact solution, using common logarithms, and a two-decimal-place approximation of each solution.

3^(x+1)=2^(1-3x)

Answer 1

\[\ln 3^{x+1} = \ln 2^{1-3x}\]
\[(x+1)ln 3 = (1-3x)ln 2\]
\[xln3+ln3=ln2-3xln2\]
\[xln3+3xln2=ln2-ln3\]
\[x(ln3+3ln2)=ln2-lnl3\]
\[x=\frac{\ln2+\ln3}{3\ln2+\ln3}\]

Answer 2

I guess I didn't think of it that way. Thanks for the help!

Answer 3

Oh, that should be ln2-ln3 in the numerator.

Answer 4

Yes, I made a typo in the last step.