Question 172156
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That's not what I got.

{{{10^(2x-1) = e^(4x-3)}}}

Let's take the natural log of both sides:

{{{ln(10^(2x-1)) = ln(e^(4x-3))}}}

Use the rules of logarithms:

{{{(2x-1)ln(10) = 4x-3}}}

Let {{{ln(10)=L}}} for ease of writing:

{{{(2x-1)L = 4x-3}}}

Reverse the factors to make it more
like standard problems:

{{{L(2x-1) = 4x-3}}}

Distribute:

{{{2Lx-L = 4x-3}}}

Get all and only x-terms on the left

{{{2Lx-4x=L-3}}}

Factor out x on the left:

{{{x(2L-4)=L-3}}}

Divide both sides by {{{(2L-4)}}}

{{{x=(L-3)/(2L-4)}}}

Calculate {{{L}}} which equals ln(10)
or 2.302585093, plug that in for L and get

{{{x=-1.152427735}}}

Edwin</pre>