What is the answer to log_{10}(2^{x+1})=3^x ?

The answer to log_{10}(2^{x+1})=3^x is No Solution for x\in\mathbb{R}

Solve for x: log_{10}(2^{x+1})=3^x

The solution is No Solution for x\in\mathbb{R}

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lo g_{10} (2^{x + 1}) = 3^{x}

No Solution for x \in R

Solution steps

lo g_{10} (2^{x + 1}) = 3^{x}

f (x) = g (x),

then

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

1 0^{l o g_{10} (2^{x + 1})} = 1 0^{3^{x}}

Simplify

1 0^{l o g_{10} (2^{x + 1})} : 2^{x + 1}

2^{x + 1} = 1 0^{3^{x}}

Solve

2^{x + 1} = 1 0^{3^{x}} :

No Solution for

x \in R

No Solution for x \in R

1 = \frac{1}{ln ( x )}

3 ln (7 x) + 7 = 19

lo g_{3} (x + 9) + 4 = 0

lo g_{3} (lo g_{5} (x)) = 2

x, y = ln (\frac{1}{2} x)

What is the answer to log_{10}(2^{x+1})=3^x ?
The answer to log_{10}(2^{x+1})=3^x is No Solution for x\in\mathbb{R}
Solve for x: log_{10}(2^{x+1})=3^x
The solution is No Solution for x\in\mathbb{R}