MHB Solving Logarithmic Equations: Expert Help with Different Bases

[lastochka]

Hello,

I am trying to solve this, but the bases are different and I am not sure how to proceed with it...
Solve the following equation. If necessary, enter your answer as an expression involving natural logarithms or as a decimal approximation that is correct to at least four decimal places.

${2}^{2x+19}$=${3}^{x-31}$

Please, help!
Thank you

 

[MarkFL]

Try taking the natural log of both sides, then apply the rule:

$$\log_a\left(b^c\right)=c\log_a(b)$$

What do you get...can you now solve for $x$?

 

[lastochka]

Thank you for the answer, but I am still not sure...
Here is what I did
$\ln\left({{2}^{2x+19}}\right)$=$\ln\left({{3}^{x-31}}\right)$

2x+19$\ln\left({2}\right)$=x-31$\ln\left({3}\right)$

x=$\frac{-31$\ln\left({3}\right)}{-19$\ln\left({2}\right)}$

x=2.58599

But the answer is incorrect.
Please, let me know what am I doing wrong

 

[MarkFL]

Okay, after applying the log rule I posted, you should have:

$$(2x+19)\ln(2)=(x-31)\ln(3)$$

Upon distribution of the logs, you then get:

$$2x\ln(2)+19\ln(2)=x\ln(3)-31\ln(3)$$

Now try solving for $x$...:D

 

[lastochka]

Yes, thank you! I just realized that I didn't do another step...must be tired...
Here what I have so far
2lnx-xln3=-31ln3-19ln2

x(2ln2) - (ln3)=-31ln3-19ln2

x=$\frac{-31ln3-19ln2}{2ln2-ln3}$

x=-164.163

I checked, the answer is right.
MarkFL, thank you so much for help!