Question 1134204
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The solution using logarithms is valid, but far more work than is needed.<br>
Solve the problem by putting everything in terms of prime factors.<br>
{{{6^(3x-4)*2^(2x+3) = 648(48^x)}}}
{{{2^(3x-4)*3^(3x-4)*2^(2x+3) = (2^3)*(3^4)*((2^4)*3)^x}}}
{{{2^(3x-4)*3^(3x-4)*2^(2x+3) = (2^3)*(3^4)*(2^(4x))*(3^x)}}}
{{{2^(5x-1)*3^(3x-4) = 2^(4x+3)*3^(x+4)}}}<br>
Equate the exponents of each prime factor:<br>
{{{5x-1 = 4x+3}}}  -->  x = 4<br>
and<br>
{{{3x-4 = x+4}}}  -->  x = 4<br>