# logarithm question

• Dec 5th 2011, 05:44 AM
zResistance
logarithm question
I'm bad at maths so how do you do this question?

(3x)^lg(3) = (4x)^lg(4)

Find the exact value of 'x'

• Dec 5th 2011, 06:47 AM
princeps
Re: logarithm question
Quote:

Originally Posted by zResistance
I'm bad at maths so how do you do this question?

(3x)^lg(3) = (4x)^lg(4)

Find the exact value of 'x'

$\log_3{(3x)^{\ln(3)}}=\log_3{(4x)^{\ln(4)}} \Rightarrow \ln(3)(1+\log_3{x})=\ln(4)(\log_3{4}+\log_3{x})$

Now solve for $\log_3{x}$ . Final result is : $x=\frac{1}{12}$
• Dec 5th 2011, 07:27 AM
zResistance
Re: logarithm question
Thanks ALOT:D , greatly appreaciated for your help
• Dec 5th 2011, 10:42 AM
Soroban
Re: logarithm question
Hello, zResistance!

Even if you're good at maths, this problem is a killer!

Here's a slightly different approach.

Quote:

$\text{Solve for }x\!:\;\;(3x)^{\ln 3}\:=\:(4x)^{\ln4}$

Take logs: . $\ln\left[(3x)^{\ln 3}\right] \:=\:\ln\left[(4x)^{\ln4}\right]$

n . . . . . . . . $\ln3 \ln(3x) \:=\:\ln4 \ln(4x)$

a . . . . $\ln 3\big[\ln 3 + \ln x\big] \:=\: \ln4\big[\ln4 + \ln x\big]$

. . . . . $(\ln 3)^2 + \ln3\ln x \:=\:(\ln4)^2 + \ln4\ln x$

. . . . $\ln4\ln x - \ln3\ln x \:=\:-(\ln4)^2 + (\ln3)^2$

x . . . . $(\ln4 - \ln3)\ln x \:=\:-\big[(\ln4)^2 - (\ln3)^2\big]$

x . . . . . . . . . . . . $\ln x \:=\:-\frac{(\ln4 - \ln3)(\ln4+\ln3)}{\ln4-\ln3}$

x . . . . . . . . . . . . $\ln x \:=\:-(\ln 4+\ln3) \:=\:-\ln(3\cdot4)$

x . . . . . . . . . . . . $\ln x \:=\:-\ln12 \:=\:\ln(12)^{-1}$

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

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