Logarithms

**Drdj** · March 29th 2011, 10:25 PM

Solve for exact value of x for (3x)^lg3 = (4x)^lg 4
[Ans = 1/12]

I took lg on both sides and rearranged to get
lg 3x / lg 4x = lg 4 / lg 3
lg 3x / lg 4x = 1.262 (evaluate)
lg 3x = 1.262 lg 4x (Cross-multiply)
lg 3x - lg 4x^1.262 = 0
lg (3x/4x^1.262) = 10^0
3x = 4x^1.262
3/4 = x^1.262 / x
Solving, x = 3/4^(1/0.262) = 0.334

Please advise where I went wrong with my working, coz my answers' not right!
Thanks !

**Sambit** · March 29th 2011, 11:34 PM

I think you have taken the log with base 10. The answer 1/12 is obtained when 'e' is used as the base. Perhaps you have done it correct otherwise. I am giving here another approach of solution:
Rewrite the given equation as $3^{log(3)} x^{log3} = 4^{log4} x^{log4}$
=> $\frac{3^{ln3}}{4^{ln4}} = x^{ln 4 - ln 3}$
=> $\frac{3^{ln3}}{4^{ln4}} = x^{ln \frac{4}{3}}$
=> $0.489=x^{0.288}$
=> $x=e^{\frac{ln 0.489}{0.288}}$
=> $x=0.0833 =\frac{1}{12}$

**Soroban** · March 30th 2011, 05:32 AM

Hello, Drdj!

If we are very careful, we can solve it without a calculator.

$\text{Solve for }x\!:\;\;(3x)^{\log3} \:=\: (4x)^{\log4}$

$\text{Answer: }\:\frac{1}{12}$

Take logs (base 10): . $\log(3x)^{\log 3} \:=\:\log(4x)^{\log4}$

. . . . . . . . . . . . . . $\log3\!\cdot\!\log(3x) \:=\:\log4\!\cdot\!\log(4x)$

. . . . . . . . . . $\log3\left[\log3 + \log x\right] \:=\:\log4\left[\log4 + \log x\right]$

. - - . . . . . . $(\log3)^2 + \log3\!\cdot\!\log x \:=\: (\log4)^2 + \log4\!\cdot\!\log x$

. . = . . . $\log4\!\cdot\!\log x - \log3\!\cdot\!\log x \:=\:(\log3)^2 - (\log4)^2$

. . . . . . . . . . $(\log4 - \log 3)\log x \:=\:-\left[(\log 4)^2 - (\log 3)^2\right]$

. . . . . . . . . . . . . . . . . . . . $\log x \:=\:-\dfrac{(\log 4 - \log 3)(\log 4 + \log 3)}{\log 4 - \log 3}$

. . . . . . . . . . . . . . . . . . . . $\log x \:=\:-\left(\log 4+\log3)$

. . . . . . . . . . . . . . . . . . . . $\log x \:=\:-\log(4\!\cdot\!3)$

. . . . . . . . . . . . . . . . . . . . $\log x \:=\:-\log12$

. . . . . . . . . . . . . . . . . . . . $\log x \:=\:\log(12^{-1})$

. . . . . . . . . . . . . . . . . . . . $\log x \:=\:\log\left(\frac{1}{12}\right)$

. . . . . . . . . . . .Therefore: . . . $x \:=\:\frac{1}{12}$

**Drdj** · March 30th 2011, 06:41 AM

Thanks for the no-calculator method !!

**Drdj** · March 30th 2011, 06:41 AM

Thanks for the alternative method!

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