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

Question

4357 views around the world

You can reuse this answer
Creative Commons License

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#