3^2x-1=5^x I am having trouble solving this since the bases are not the same and I am unclear how to use Ln or Log to solve even though I know I...

Thursday, November 5, 2009

3^2x-1=5^x I am having trouble solving this since the bases are not the same and I am unclear how to use Ln or Log to solve even though I know I...

Hello!

I think your equation is

$\displaystyle{{3}}^{{{2}{x}-{1}}}={{5}}^{{x}},$

not

$\displaystyle{{3}}^{{{2}{x}}}-{1}={{5}}^{{x}}.$

For the first option yes, apply $\displaystyle{\ln{}}$ to both sides (they are both positive). It is known (and not hard) that $\displaystyle{\ln{{\left({{a}}^{{b}}\right)}}}={b}\cdot{\ln{{\left({a}\right)}}}.$

Here we obtain $\displaystyle{\left({2}{x}-{1}\right)}\cdot{\ln{{\left({3}\right)}}}={x}\cdot{\ln{{\left({5}\right)}}}.$

It is a linear equation for x, group like terms:

$\displaystyle{x}\cdot{\left({2}{\ln{{\left({3}\right)}}}-{\ln{{\left({5}\right)}}}\right)}={\ln{{\left({3}\right)}}},$ so

$\displaystyle{x}=\frac{{{\ln{{\left({3}\right)}}}}}{{{2}{\ln{{\left({3}\right)}}}-{\ln{{\left({5}\right)}}}}}\approx{1.869}.$

This is the answer for the first option.

I believe the second option cannot be solved exactly, although it has exactly one root and it is in (0, 1).

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Thursday, November 5, 2009

3^2x-1=5^x I am having trouble solving this since the bases are not the same and I am unclear how to use Ln or Log to solve even though I know I...

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