Given that (3^(x+2))*(5^(x-1)) = 15^(2x), show that 15^x = 9/5.
I managed to reach: 15^x= [3*sqrt(5)*sqrt(3^x)*sqrt(5^x)]/5, but can't reach 9/5.
Are you sure you copied the question correctly? as:
$\displaystyle 15x=\frac{9}{5}$
$\displaystyle \therefore x=\frac{3}{25}$
So,
$\displaystyle \left (3\times\frac{3}{25}+2\right )\left (5\times \frac{3}{25}-1\right )$ must equal $\displaystyle 152\times\frac{3}{25}$
but this gives $\displaystyle -\frac{118}{125}=\frac{456}{25}$, which is false.
$\displaystyle \displaystyle ({3^{x+2}})(5^{x-1}) = 15^{2x}$ $\displaystyle \displaystyle \Rightarrow (3^{x+2})(5^{x-1}) = 15^{x}\cdot 15^{x}$ $\displaystyle \displaystyle \Rightarrow \frac{3^x\cdot 3^2 \cdot 5^{x}\cdot 5^{-1}}{15^x} = 15^{x}$ $\displaystyle \displaystyle\Rightarrow \frac{9}{5}\left(\frac{3^{x}\cdot 5^{x}}{15^x}\right) = 15^{x}$.
$\displaystyle 3^{x + 2}\cdot 5^{x - 1} = 15^{2x}$
$\displaystyle \ln{\left(3^{x + 2}\cdot 5^{x - 1}\right)} = \ln{\left(15^{2x}\right)}$
$\displaystyle \ln{\left(3^{x + 2}\right)} + \ln{\left(5^{x - 1}\right)} = 2x\ln{(15)}$
$\displaystyle (x + 2)\ln{3} + (x - 1)\ln{5} = 2x(\ln{3} + \ln{5})$
$\displaystyle x\ln{3} + 2\ln{3} + x\ln{5} - \ln{5} = 2x\ln{3} + 2x\ln{5}$
$\displaystyle 2\ln{3} - \ln{5} = x\ln{3} + x\ln{5}$
$\displaystyle 2\ln{3} - \ln{5} = x(\ln{3} + \ln{5})$
$\displaystyle \frac{2\ln{3} - \ln{5}}{\ln{3} + \ln{5}} = x$.
I think I misread your question earlier. Apologies. The answer is because taking the square root of both sides is not going to take you anywhere.
There is no method of identifying which technique to use what time; it's just insights that one gains from exposure and experience by doing lots of this.