What is the answer to solvefor x,(e^{2x+1})/5 =yz ?

The answer to solvefor x,(e^{2x+1})/5 =yz is x=(ln(5yz)-1)/2

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solvefor x,(e^{2x+1})/5 =yz

solve for

x, \frac{e ^{2 x + 1}}{5} = yz

x = \frac{ln ( 5 yz ) - 1}{2}

Solution steps

\frac{e ^{2 x + 1}}{5} = yz

Multiply both sides by

5

\frac{e ^{2 x + 1}}{5} \cdot 5 = yz \cdot 5

Simplify

e^{2 x + 1} = yz \cdot 5

Apply exponent rules

2 x + 1 = ln (yz \cdot 5)

Solve

2 x + 1 = ln (yz \cdot 5) : x = \frac{ln ( 5 yz ) - 1}{2}

x = \frac{ln ( 5 yz ) - 1}{2}

15 e^{5 x} = 5

62 \cdot 1 0^{x} = 40

7 e^{3 n - 5} + 5 = - 89

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

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

What is the answer to solvefor x,(e^{2x+1})/5 =yz ?
The answer to solvefor x,(e^{2x+1})/5 =yz is x=(ln(5yz)-1)/2