Question

If 3^1-x +3^1+x = 90, Find the value of x

Answer 1

`x=\log_3{(15+ 4√(14))}\approx3.0949\text{ or } \\x=\log_3{(15- 4√(14))}\approx-3.0949`

Explanation:

So we have the equation:

`3^(1-x)+3^(1+x)=90`

First, note that this is the same as:

`3\cdot 3^(-x)+3\cdot 3^x=90`

Factor out a 3:

`3(3^(-x)+3^(x))=90`

Divide both sides by 3:

`3^(-x)+3^x=30`

Change the negative exponent into a fraction:

`(1)/(3^x)+3^x=30`

Multiply both sides by 3^x:

`3^x((1)/(3^x)+3^x)=(30)(3^x)`

Distribute:

`(1)+(3^x)^2=30(3^x)`

Let u equal 3^x. So:

`1+u^2=30u`

Subtract 30u from both sides:

`u^2-30u+1=0`

This is a quadratic. Solve using the quadratic formula. A is 1, b is -30, and c is 1. So:

`u=(-b\pm √(b^2-4ac))/(2a)`

Substitute:

`u=(-(-30)\pm √((-30)^2-4(1)(1)))/(2(1))`

Simplify:

`u=(30\pm √(900-4))/(2)`

Subtract:

`u=(30\pm √(896))/(2)`

Simplify:

`u=(30\pm 8√(14))/(2)`

Simplify:

`u=15\pm 4√(14)`

Substitute back u:

`3^x=15\pm 4√(14)`

Take the log to base 3 of both sides:

`\log_3{3^x}=\log_3{15\pm 4√(14)}`

The left side cancels:

`x=\log_3{15\pm 4√(14)}`

So, our solutions are:

`x=\log_3{(15+ 4√(14))}\approx3.0949\text{ or } \\x=\log_3{(15- 4√(14))}\approx-3.0949`

And we're done!