What is the answer to (e^{3x+1})/(e^{x-1)}=2 ?

The answer to (e^{3x+1})/(e^{x-1)}=2 is x=(ln(2)-2)/2

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(e^{3x+1})/(e^{x-1)}=2

\frac{e ^{3 x + 1}}{e ^{x - 1}} = 2

x = \frac{ln ( 2 ) - 2}{2}

Solution steps

\frac{e ^{3 x + 1}}{e ^{x - 1}} = 2

Apply exponent rules

2 x + 2 = ln (2)

Solve

2 x + 2 = ln (2) : x = \frac{ln ( 2 ) - 2}{2}

x = \frac{ln ( 2 ) - 2}{2}

Verify Solutions:

x = \frac{ln ( 2 ) - 2}{2}

True

The solution is

x = \frac{ln ( 2 ) - 2}{2}

5^{x} = 21

2^{x^{2}} = 8^{2 x - 3}

(\frac{1}{3})^{x + 2} = 3^{x - 1}

2^{x^{2} - x - 3} = \frac{1}{2}

5 \cdot 2^{3 x} = 15000

What is the answer to (e^{3x+1})/(e^{x-1)}=2 ?
The answer to (e^{3x+1})/(e^{x-1)}=2 is x=(ln(2)-2)/2