# Solve for x equation 7^x+7^-x=1?

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### 2 Answers

If your question was `7^x-7^(-x)=1`

Which would make more sense, We get

`(7^x)^2-7^x-1=0`

a=1, b=-1, c=-1 gives us using the quadradic formula

This gives `7^x=(1+-sqrt(1+4))/2=(1+-sqrt(5))/2`

Since 1-sqrt(5) is negative and 7^x is always positive, we get

`log_7(7^x)=x=log_7((1+sqrt(5))/2)=(log_10((1+sqrt(5))/2))/(log_10(7))~~`0.247293959 227

You need to use the negative power property of exponentials such that: `7^-x = 1/(7^x)`

Write the equation such that:

`7^x + 1/(7^x) = 1`

Bringing all terms to a common denominator yields:

`7^(2x) + 1 = 7^x`

You should come up with the substitution `7^x = y`

`` `y^2 + 1 = y`

Subtracting y both sides yields:

`y^2 - y + 1 = 0`

You need to use quadratic formula:

`y_(1,2) = (1+-sqrt(1-4))/2 =gt y_(1,2) = (1+-isqrt3)/2`

`` `7^x = (1+-isqrt3)/2`

**Since the result to an exponential equation is a real number, hence the equation has no solutions.**