Answer:
[tex]x=\frac{\ln(5)+1}{\ln(2)-1}\approx-8.5039[/tex]
Step-by-step explanation:
So we have the equation:
[tex]2^x=5e^{x+1}[/tex]
Let's take the natural log of both sides:
[tex]\ln(2^x)=\ln(5e^{x+1})[/tex]
On the left, we can move the x to the front:
[tex]x\ln(2)=\ln(5e^{x+1})[/tex]
On the right, we can separate the logs:
[tex]x\ln(2)=\ln(5)+\ln(e^{x+1})[/tex]
For the second term on the right, move all the exponent stuff to the front:
[tex]x\ln(2)=\ln(5)+(x+1)\ln(e)[/tex]
The natural log of e is 1. So:
[tex]x\ln(2)=\ln(5)+(x+1)(1)[/tex]
Simplify:
[tex]x\ln(2)=\ln(5)+x+1[/tex]
Subtract x from both sides:
[tex]x\ln(2)-x=\ln(5)+1[/tex]
Factor out an x from the left:
[tex]x(\ln(2)-1)=\ln(5)+1[/tex]
Divide both sides by the equation in the factor:
[tex]x=\frac{\ln(5)+1}{\ln(2)-1}[/tex]
So, our answer is:
[tex]x=\frac{\ln(5)+1}{\ln(2)-1}\approx-8.5039[/tex]
And we're done!
