How do I use logarithms to solve for $x$ if $7^(2x-1)=316$ ?

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Answer 1 · 2014-09-11T01:15:30

Please take a look at this video that gives step-by-step instructions to solve a very similar problem.

$7^(2x-1)=316$

Take the natural log of both sides of the equation

$ln(7^(2x-1))=ln(316)$

Use the properties of logarithms to move the exponent

$(2x-1)*ln(7)=ln(316)$

Use basic algebra skills to solve

$((2x-1)*ln(7))/ln(7)=ln(316)/ln(7)$

$2x-1=ln(316)/ln(7)$

$2x=ln(316)/ln(7)+1$

$(2x)/2=(ln(316)/ln(7)+1)/2$

$x=(ln(316)/ln(7)+1)/2$

$x=1.978933191$

How do I use logarithms to solve for xx if 7^(2x-1)=3167^(2x-1)=316?