# Logarithms!

• May 22nd 2008, 06:20 PM
andymac
Logarithms!
I need some help solving the following:

a) (2)*3^x = (7)*5^x

b) 12^x = (4)*8^(2x)

c)(12)*6^(2x-1) = 11^(x+3)

d) (7)*0.43^(2x) = (9)*6^(-x)

Thanks.
• May 22nd 2008, 06:28 PM
Mathstud28
Quote:

Originally Posted by andymac
I need some help solving the following:

a) (2)*3^x = (7)*5^x

b) 12^x = (4)*8^(2x)

c)(12)*6^(2x-1) = 11^(x+3)

d) (7)*0.43^(2x) = (9)*6^(-x)

Thanks.

Two methods for these. I will show you both for the first problem and you can apply it to the rest

Method one

$2\cdot{3^x}=7\cdot{5^{x}}$

taking the natural log of both sides gives

$ln(2\cdot{3^{x}})=\ln(2)+\ln(3)x=\ln(7\cdot{5^x})= \ln(7)+\ln(5)x$

Algebraically manipulating gives

$ln(3)x-\ln(5)x=x(\ln(3)-\ln(5))=x\ln\bigg(\frac{3}{5}\bigg)=\ln(7)-\ln(2)=\ln\bigg(\frac{7}{2}\bigg)$

so $x=\frac{\ln\bigg(\frac{7}{2}\bigg)}{\ln\bigg(\frac {3}{5}\bigg)}$

Method two

$2\cdot{3^x}=7\cdot{5^x}$

through manipulation we get

$\frac{3^x}{5^x}=\bigg(\frac{3}{5}\bigg)^x=\frac{7} {2}$

taking the natural log of both sides gives

$x\ln\bigg(\frac{3}{5}\bigg)=\ln\bigg(\frac{7}{2}\b igg)\Rightarrow{x=\frac{\ln\bigg(\frac{7}{2}\bigg) }{\ln\bigg(\frac{3}{5}\bigg)}}$