What is the answer to log_{9}(3^{x+1})=x^2 ?

The answer to log_{9}(3^{x+1})=x^2 is x=1,x=-1/2

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log_{9}(3^{x+1})=x^2

lo g_{9} (3^{x + 1}) = x^{2}

x = 1, x = - \frac{1}{2}

Solution steps

lo g_{9} (3^{x + 1}) = x^{2}

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

9^{l o g_{9} (3^{x + 1})} = 9^{x^{2}}

Simplify

9^{l o g_{9} (3^{x + 1})} : 3^{x + 1}

3^{x + 1} = 9^{x^{2}}

Solve

3^{x + 1} = 9^{x^{2}} : x = 1, x = - \frac{1}{2}

x = 1, x = - \frac{1}{2}

Verify Solutions:

x = 1

True

, x = - \frac{1}{2}

True

The solutions are

x = 1, x = - \frac{1}{2}

2 x = e^{2} ln (x)

lo g_{10} (x) = 7.6348

5 = lo g_{x} (32)

lo g_{7} (x) = 21

0.9 = ln (y)

What is the answer to log_{9}(3^{x+1})=x^2 ?
The answer to log_{9}(3^{x+1})=x^2 is x=1,x=-1/2
Solve for x: log_{9}(3^{x+1})=x^2
The solution is x=1,x=-1/2