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

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.

  • $\begingroup$ 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 $\log_b(a) = \frac{\ln(a)}{\ln(b)}$. $\endgroup$ May 13, 2016 at 8:29
  • $\begingroup$ "the bases are all different": then convert to a common base. $\endgroup$
    – user65203
    May 13, 2016 at 9:06

1 Answer 1


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.

  • $\begingroup$ Yes that's what I meant. I can understand upto the 3rd step. Now should I approach by taking the formula logA/B? $\endgroup$
    – Ritwika
    May 13, 2016 at 8:37
  • $\begingroup$ @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... $\endgroup$
    – DonAntonio
    May 13, 2016 at 8:38
  • $\begingroup$ Actually I am not being to change the base after your 3rd step. Can you please help me? $\endgroup$
    – Ritwika
    May 13, 2016 at 8:43
  • $\begingroup$ @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 $\;\sqrt x=x^{1/2}\;$ . $\endgroup$
    – DonAntonio
    May 13, 2016 at 8:44
  • 1
    $\begingroup$ @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. $\endgroup$
    – DonAntonio
    May 13, 2016 at 12:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.