# Solve for x: $4\log_{x/2}(\sqrt{x}) + 2 \log_{4x} (x^2) = 3 \log_{2x} (x^3)$

$$4\log_{x/2}(\sqrt{x}) + 2 \log_{4x} (x^2) = 3 \log_{2x} (x^3)$$

This is a different type of equation. Our school has not taught this type yet. But this came in our exams. Can someone please help? I don't understand the bases are all different. How can I proceed? Please I need a step wise solution. Thank you.

• Do you mean $2\log_{4x}(x^2)$ by 2 log(x^2) base(4x)? I don't even know if it's a allowed to have a non-constant variable as the base, but you may be able to use $\log_b(a) = \frac{\ln(a)}{\ln(b)}$. May 13 '16 at 8:29
• "the bases are all different": then convert to a common base.
– user65203
May 13 '16 at 9:06

Let me see if I guessed correctly what you wrote:

$$4\log_{x/2}\sqrt x+2\log_{4x}x^2=3\log_{2x}x^3\iff2\log_{x/2}x+4\log_{4x}x=9\log_{2x}x\implies$$

$$2\frac{\log x}{\log\frac x2}+4\frac{\log x}{\log 4x}=9\frac{\log x}{\log 2x}$$

where $\;log\;$ above can be at any base you want (though in higher mathematics it is usually taken to be $\;\log_e\;$)

Now you could assume $\;x\neq 1\;$ (otherwise the exercise is very boring) and thus divide through the whole last equation by $\;\log x\;$ , and then use other properties of logarithms, say like $\;\log AB=\log A+\log B\;$ , or $\;\log\frac AB=\log A-\log B\;$ , etc.

Also make sure you understand and can justify the first steps shown above.

• Yes that's what I meant. I can understand upto the 3rd step. Now should I approach by taking the formula logA/B? May 13 '16 at 8:37
• @Ritwika If you mean the jump from the first to the second line of equations is hard: use what Maximilian proposes in his comment: change of base in logarithms. That is what I used there... May 13 '16 at 8:38
• Actually I am not being to change the base after your 3rd step. Can you please help me? May 13 '16 at 8:43
• @Ritwika Do you mean, for example, that $\;4\log_{x/2}\sqrt x=2\log_{x/2}x\;$ ? That only uses the property $\;\log x^n=n\log x\;$ and also the trivial $\;\sqrt x=x^{1/2}\;$ . May 13 '16 at 8:44
• @Ritwika Whatever you do (mathematically sound, of course) with one side of an equality you must do exactly the same on the side...always. I think that dividing by $\;\log x\neq0\iff x\neq1\;$ in the last equality can help to make things a little simpler. May 13 '16 at 12:38