Question 153490
I'm assuming this is your equation (not sure w/o parentheses),
{{{e^(5x-3)=10}}}
{{{5x-3=ln(10)}}}Natural log of boths sides.
{{{5x=ln(10)+3}}}
{{{x=(ln(10)+3)/5}}}
or approximately,
{{{x=1.061}}}
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If your equation was,
{{{e^(5x)-3=10}}}
{{{e^(5x)=13}}}
{{{5x=ln(13)}}}
{{{x=ln(13)/5}}}
{{{x=0.513}}}