Question 1095930
?  "same bases"?


{{{ln(5^(2x+1))=ln(2^(x-2))}}}

{{{(2x+1)ln(5)=(x-2)(ln(2))}}}

You can use a handheld calculator or some other source for the ln(5) and the ln(2).  I will simply represent them with constant variables {{{k=ln(5)}}} and {{{h=ln(2)}}}.

{{{(2x+1)k=(x-2)h}}}

{{{2kx+k=hx-2h}}}

{{{2kx-hx=-k-2h}}}

{{{(2k-h)x=-k-2h}}}

{{{highlight_green(x=(-k-2h)/(2k-h))}}}-----just substitute and use your calculator to finish.