How to solve $\log_{4}(\sqrt{x^{4/3}})+3\log_{x}(16x)=7?$
I've tried everything from brute force to doing base change but nothing works. I was wondering what was the best way in solving this?
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Consider changing the logarithmic base to a natural log, ie $$\log_{b}(x) = \frac{\ln(x)}{\ln(b)}$$ which then leads to \begin{align} \log_{4}(x^{2/3}) + 3 \, \log_{x}(16 \, x) &= 7 \\ \frac{\ln(x)}{3 \, \ln(2)} + \frac{3 \, \ln(16 \, x)}{\ln(x)} &= 7 \\ \frac{\ln(x)}{3 \, \ln(2)} + \frac{12 \, \ln(2) + 3 \, \ln(x)}{\ln(x)} &= 7 \\ \ln^{2}(x) - 12 \, \ln(2) \, \ln(x) + 36 \, \ln^{2}(2) &= 0 \\ \left(\ln(x) - 6 \, \ln(2) \right)^2 &= 0 \end{align} for which it is obtained $\ln(x) = 6 \, \ln(2) = \ln(2^{6})$ or $x = 2^{6} = 64$.
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\begin{align} \log_{4}(\sqrt{x^{4/3}})+3\log_{x}(16x)&=7\\ \left(\frac{1}{2}\right)\left(\frac{4}{3}\right)\log_{4}(x)+3\log_{x}(4^2)+3\log_{x}(x)&=7\\ \frac{2}{3}\log_{4}(x)+\frac{6}{\log_4x}+3&=7\\ (\log_4x)^2-6\log_4x+9&=0\\ \log_4x&=3\\ x&=64 \end{align}
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$\begingroup$ I understand everything except on how $3log_{x}(4^2)$ became $\frac{6}{log_4x}$? $\endgroup$ – John Rawls Jun 15 '17 at 1:37
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1$\begingroup$ $3\log_x(4^2)=6\log_x4$. $\log_x4=\frac{\log_44}{\log_4x}=\frac{1}{\log_4x}$ by the change of base formula. $\endgroup$ – CY Aries Jun 15 '17 at 1:38
