Solve for: `x :(2x)^((log)_b2)=(3x)^((log)_b3)` .

Explanation

` (2x)^(log_(b) 2) = (3x)^(log_(b)3)`
` rArr log_(b) 2 [ log 2 + log x ]= log_(b) 3 [log3 + log x]`
` or (log_(b)2)(log 2) - log_(b)3*log 3 = (log_(b)3-log_(b) 2) log x`
` or (log 2)/(log b) * log 2 - (log 3)/(log b) * log 3 = ((log 3)/(logb)-(log 2)/(logb)) log x`
` or ((log 2)^(2) - (log 3)^(2))/(log b) = ((log 3 - log 2)/(log b)) log x `
` or log x =- (log 3 + log 2) = log (6)^(-1)`
` or x = 1//6`