Answer:
[tex]x=\log_3{(15+ 4\sqrt{14})}\approx3.0949\text{ or } \\x=\log_3{(15- 4\sqrt{14})}\approx-3.0949[/tex]
Step-by-step explanation:
So we have the equation:
[tex]3^{1-x}+3^{1+x}=90[/tex]
First, note that this is the same as:
[tex]3\cdot 3^{-x}+3\cdot 3^x=90[/tex]
Factor out a 3:
[tex]3(3^{-x}+3^{x})=90[/tex]
Divide both sides by 3:
[tex]3^{-x}+3^x=30[/tex]
Change the negative exponent into a fraction:
[tex]\frac{1}{3^x}+3^x=30[/tex]
Multiply both sides by 3^x:
[tex]3^x(\frac{1}{3^x}+3^x)=(30)(3^x)[/tex]
Distribute:
[tex](1)+(3^x)^2=30(3^x)[/tex]
Let u equal 3^x. So:
[tex]1+u^2=30u[/tex]
Subtract 30u from both sides:
[tex]u^2-30u+1=0[/tex]
This is a quadratic. Solve using the quadratic formula. A is 1, b is -30, and c is 1. So:
[tex]u=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Substitute:
[tex]u=\frac{-(-30)\pm \sqrt{(-30)^2-4(1)(1)}}{2(1)}[/tex]
Simplify:
[tex]u=\frac{30\pm \sqrt{900-4}}{2}[/tex]
Subtract:
[tex]u=\frac{30\pm \sqrt{896}}{2}[/tex]
Simplify:
[tex]u=\frac{30\pm 8\sqrt{14}}{2}[/tex]
Simplify:
[tex]u=15\pm 4\sqrt{14}[/tex]
Substitute back u:
[tex]3^x=15\pm 4\sqrt{14}[/tex]
Take the log to base 3 of both sides:
[tex]\log_3{3^x}=\log_3{15\pm 4\sqrt{14}}[/tex]
The left side cancels:
[tex]x=\log_3{15\pm 4\sqrt{14}}[/tex]
So, our solutions are:
[tex]x=\log_3{(15+ 4\sqrt{14})}\approx3.0949\text{ or } \\x=\log_3{(15- 4\sqrt{14})}\approx-3.0949[/tex]
And we're done!
