Having trouble solving this simple logarithm problem:
$$\log_3(x^3) - 4\log_9(x) - 5\log_{27}(x^{1/2}) = \log_9(4)$$
I’ve been stuck as I when I solve it I get an answer of $2$, when the actual solution seems to be $x=64$?
Whilst editing and trying to explain my working I found the error I did. Solution: $$ \begin{align} \log_3(x^3) - 4\log_9(x) - 5\log_{27}(x^{1/2}) = \log_9(4)& \\ \log_3(x^3) - {\log_3{x^4}}/{\log_3(9)} - {\log_3(x^{1/2})^5}/{\log_3(27)} = {\log_3(4)}/{\log_3(9)}& \\ \log_3(x^3) - {\log_3{x^4}}/{2} - {\log_3(x^{1/2})^5}/{3} = {2\log_3(2)}/{2}& \\ 6\log_3(x^3)/{6} - {3\log_3{x^4}}/{6} - {2\log_3(x^{1/2})^5}/{6} = {\log_3(2)}& \\ 6\log_3(x^3) - {3\log_3{x^4}} - {2\log_3(x^{1/2})^5} = {\log_3(64)}& \\ \log_3(x^3)^6 - {\log_3(x^4)^3} - {\log_3(x^{1/2})^{10}} = {\log_3(64)}& \\ \log_3(x^6)- {\log_3(x^{5})} = {\log_3(64)}& \\ \log_3{x}= {\log_3(64)}& \end{align} $$ And since $\log_c(a) = \log_c(b)$, $a=b$, the solution is
$$x=64$$
Apologies if there is a nicer solution, I tried my best!
By the exponent rule for logarithms, we have \begin{align*} 3\log _3 x - 4\log _9 x - \frac{5}{2}\log _{27} x = \log _9 4 \end{align*} and by change of base rule we have \begin{align*} 3\log _3 x - 4\cdot\frac{\log _3 x}{\log _3 9} - \frac{5}{2}\cdot\frac{\log _3 x}{\log _3 27} = \frac{\log _3 4}{\log _3 9}\\ 3\log _3 x - 4\cdot\frac{\log _3 x}{2} - \frac{5}{2}\cdot\frac{\log _3 x}{3} = \frac{\log _3 4}{2}\text{.} \end{align*} Simplifying yields \begin{align*} \log _3 x &= 3\cdot\log _3 4\\ \log _3 x &= \log _3 (4^3)\\ x &= 4^3=64\text{.} \end{align*}