logs

[IssyAspen]

Hello

I am working through a problem and cannot see how they go from one step to the next. Could someone please help.

The question is:

Solve the equation

\(\displaystyle (3^x)(4^{2x+1}) = 6^{x+2}\)

giving your answer in the form \(\displaystyle \dfrac{\ln {a}}{\ln {b}}\)

I followed the working up to here:

\(\displaystyle x\ln3 + (2x+1)\ln4 = (x+2)\ln6\)


Bu then don’t see what they’ve done to get this:

\(\displaystyle \Rightarrow x(\ln{3} + 2\ln{4} -\ln {6} = 2\ln{6} - \ln{4}\)


Thank you.

 

[srmichael]

Hello

I am working through a problem and cannot see how they go from one step to the next. Could someone please help.

The question is:

Solve the equation

\(\displaystyle (3^x)(4^{2x+1}) = 6^{x+2}\)

giving your answer in the form \(\displaystyle \dfrac{\ln {a}}{\ln {b}}\)

I followed the working up to here:

\(\displaystyle x\ln3 + (2x+1)\ln4 = (x+2)\ln6\)


Bu then don’t see what they’ve done to get this:

\(\displaystyle \Rightarrow x(\ln{3} + 2\ln{4} -\ln {6} = 2\ln{6} - \ln{4}\)


Thank you.

They distributed the ln4 and the ln6, combined like terms, factored out the "x" and brought the constants, 2ln6 and ln4 to the right hand side of the equal sign.

 

[pka]

\(\displaystyle \Rightarrow x(\ln{3} + 2\ln{4} -\ln {6} )= 2\ln{6} - \ln{4}\)

\(\displaystyle \ln{3} + 2\ln{4} -\ln {6}=\ln\left(\dfrac{3\cdot 16}{6}\right)\)

 

[IssyAspen]

Hello srmichael

They distributed the ln4 and the ln6, combined like terms, factored out the "x" and brought the constants, 2ln6 and ln4 to the right hand side of the equal sign.

Thank you, I didn't know I could apply the distributive property to logs in that way. I do now.

Thank you for your reply.

 

[IssyAspen]

Hello PKA

\(\displaystyle \ln{3} + 2\ln{4} -\ln {6}=\ln\left(\dfrac{3\cdot 16}{6}\right)\)

Thank you.