Question 589689
what is the solution to this logarithm question. log3 to power of x=y=log9 to power(2x-1)
:
Assume the problem is: {{{log((3^x))}}} = {{{log((9^((2x-1))))}}} I don't know what the y is there for.
Find x
{{{log((3^x))}}} = {{{log((9^((2x-1))))}}}
we can change the 9 to 3^2, then we have
{{{log((3^x))}}} = {{{log((3^(2(2x-1))))}}}
{{{log((3^x))}}} = {{{log((3^((4x-2))))}}}
the log equiv of exponents
{{{x*log((3))}}} = {{{(4x-2)*log((3))}}}
there fore we know
x = 4x - 2
2 = 4x - x
2 = 3x
x = 2/3