# Solve the equation 10^(1+x^2) - 10^(1-x^2)=99

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### 2 Answers

We have to solve 10^(1+x^2) - 10^(1-x^2)=99

10^(1+x^2) - 10^(1-x^2)=99

=> 10*10^x^2 - 10/10^x^2 = 99

let 10^x^2 = y

=> 10y - 10/y = 99

=> 10y^2 - 99y - 10 = 0

=> 10y^2 - 100y + y - 10 = 0

=> 10y(y - 10) + 1(y - 10) = 0

=> (10y + 1)(y - 10) = 0

=> y = -1/10 and y = 10

10^x^2 = -1/10

This is not possible as 10 raised to any power gives a positive result.

10^x^2 = 10

=> x^2 = 1

=> x = 1 and x = -1

**The solutions of the equation are (-1, 1)**

We'll re-write the equation:

10*10^(x^2) - 10/10^(x^2) - 99 = 0

We'll replace 10^(x^2) by t:

10*t - 10/t - 99 = 0

10t^2 - 99t - 10 = 0

We'll apply quadratic formula:

t1 = [99+sqrt(9801+400)]/20

t1 = (99+101)/20

t1 = 10

t2 = (99-101)/20

t2 = -0.1

But 10^(x^2) = t1 <=> 10^(x^2) = 10

Since the bases are matching, we'll apply one to one rule:

x^2 = 1 <=> x1 = -1 and x2 = 1

10^(x^2) = t1 <=> 10^(x^2) =-0.1 , impossible.

**The possible solutions of the equation are: {-1 ; 1}.**