Question 132616
{{{2log4(2x)-log4(3x-5)=log3(9)}}}

Solution:

Rewrite the equation:

{{{log4(2x)^2-log4(3x-5)=log3(3)^2}}}

{{{log4((2x)^2/(3x-5))=2log3(3)}}}

{{{log4((2x)^2/(3x-5))=2}}}

Replace 2 by {{{log4(16)}}} since they are equal and we need base 4 log.

So we get

{{{log4((2x)^2/(3x-5))=log4(16)}}}

Compare the two sides, we get

{{{(2x)^2/(3x-5)=16}}}

Then multiply 3x-5 on both sides:

{{{(2x)^2=16(3x-5)}}}

{{{4x^2=16(3x-5)}}}

divide by 4 on both sides:

{{{x^2=4(3x-5)}}}

expand it and write into standard form:

{{{x^2-12x+20=0}}}

Factor left side:

(x-10)(x-2)=0;


So x=10 or x=2.

That's it!