What is the answer to sqrt(3^{2x)+7}=3^x+1 ?

The answer to sqrt(3^{2x)+7}=3^x+1 is x=1

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sqrt(3^{2x)+7}=3^x+1

3^{2 x} + 7 = 3^{x} + 1

x = 1

Solution steps

3^{2 x} + 7 = 3^{x} + 1

Apply exponent rules

(3^{x})^{2} + 7 = 3^{x} + 1

Rewrite the equation with

3^{x} = u

(u)^{2} + 7 = u + 1

Solve

u^{2} + 7 = u + 1 : u = 3

u = 3

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = 3 : x = 1

x = 1

\frac{9 ^{2 x}}{3 ^{x}} = 27

2^{x} + 2^{2 - x} - 5 = 0

2^{x^{3}} = 6 4^{x}

12 5^{x - 1} = 2 5^{x + 3}

2^{3 x - 1} = 0.125

What is the answer to sqrt(3^{2x)+7}=3^x+1 ?
The answer to sqrt(3^{2x)+7}=3^x+1 is x=1
Solve for x: sqrt(3^{2x)+7}=3^x+1
The solution is x=1