log3 2x log89 3x 1Solutionlog32x log93x 1 Use the followin

log_3 (2x) + log_89 (3x) = 1

log3(2x) +log9(3x) =1

Use the following log properties:

Now loga(X) = logX/loga

Further, logx^a = a*logx

So, log3(2x) + log(3x)/log9 = 1

log3(2x) +log(3x)/log3^2 =1

log3(2x) +log(3x)/ 2*log3 =1

So, log3(2x) + (1/2)log3(3x) =1

log3(2x) + log3(3x)^1/2 =1

log3{ 2x*(3x)^1/2} =1

log3(2*3^1/2*x^3/2 )=1

Use the exponential property: loga(x) =b

x = a^b

So, 2*3^1/2*x^3/2 = 3

squaring both sides: 4*3*x^3 = 9

x^3 = 9/12 = 3/4

x = (3/4)^1/3