2

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

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.

Ritwika
  • 79
  • 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 $\log_b(a) = \frac{\ln(a)}{\ln(b)}$. – Maximilian Gerhardt May 13 '16 at 08:29
  • "the bases are all different": then convert to a common base. –  May 13 '16 at 09:06

1 Answers1

4

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.

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
  • Yes that's what I meant. I can understand upto the 3rd step. Now should I approach by taking the formula logA/B? – Ritwika May 13 '16 at 08:37
  • @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... – DonAntonio May 13 '16 at 08:38
  • Actually I am not being to change the base after your 3rd step. Can you please help me? – Ritwika May 13 '16 at 08:43
  • @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 $;\sqrt x=x^{1/2};$ . – DonAntonio May 13 '16 at 08:44
  • The base is already changed, it's the $\log_{x/2}(\sqrt{x}) = \frac{\sqrt{x}}{\ln(x/2)}$ step. Then $\sqrt{x} = x^{1/2}$ and $\ln(a^b) = b\ln(a)$. Also other hint: $\log(x/2) = \log(x) - \log(2)$, also $\log(4x) = \log(x) + \log(4)$, you will be able to reduce everything to $\log(x) + \text{something}$. Also be caerful when dividing by $\ln(x)$, make sure $\ln(x) \neq 0$ and check that seperate case in case it it. – Maximilian Gerhardt May 13 '16 at 08:45
  • @Joanpemo Should I have to divide the last equation by log(x) on both left hand side and right hand side? – Ritwika May 13 '16 at 08:57
  • 1
    @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. – DonAntonio May 13 '16 at 12:38
  • I have solved this equation! But I am new to stackexchange so I don't know how to give codes so that I can post the answer. :( – Ritwika May 13 '16 at 14:08
  • Basically, I just changed your third equation a bit. I wrote, 2/ (log X/2 base x) + 4/ (log 4x base x) =9 /(log 2xbasex). As the bases are all x , I assumed log 2 base x=a and completed it. – Ritwika May 13 '16 at 14:29