What is the answer to x5^{2x+3}= 1/25 ?

The answer to x5^{2x+3}= 1/25 is x=(\W_{0}(\frac{2ln(5))/(3125))}{2ln(5)}

Solve for x: x5^{2x+3}= 1/25

The solution is x=(\W_{0}(\frac{2ln(5))/(3125))}{2ln(5)}

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x 5^{2 x + 3} = \frac{1}{25}

x = \frac{W _{0} ( \frac{2 l n ( 5 )}{3125} )}{2 ln ( 5 )}

Decimal Notation

x = 0.00031 \dots

Solution steps

x \cdot 5^{2 x + 3} = \frac{1}{25}

Prepare

x 5^{2 x + 3} = \frac{1}{25}

for Lambert form:

x e^{l n (5) (2 x + 3)} = \frac{1}{25}

Rewrite the equation with

2 x ln (5) = u

and

x = \frac{u}{2 ln ( 5 )}

(\frac{u}{2 ln ( 5 )}) e^{l n (5) (2 (\frac{u}{2 l n ( 5 )}) + 3)} = \frac{1}{25}

Simplify

\frac{e ^{u + 3 l n (5)} u}{2 ln ( 5 )} = \frac{1}{25}

Rewrite

\frac{e ^{u + 3 l n (5)} u}{2 ln ( 5 )} = \frac{1}{25}

in Lambert form:

e^{u} u = \frac{2 ln ( 5 )}{3125}

Solve

e^{u} u = \frac{2 ln ( 5 )}{3125} : u = W_{0} (\frac{2 ln ( 5 )}{3125})

u = W_{0} (\frac{2 ln ( 5 )}{3125})

Substitute back

u = 2 x ln (5),

solve for

x

Solve

2 x ln (5) = W_{0} (\frac{2 ln ( 5 )}{3125}) : x = \frac{W _{0} ( \frac{2 l n ( 5 )}{3125} )}{2 ln ( 5 )}

x = \frac{W _{0} ( \frac{2 l n ( 5 )}{3125} )}{2 ln ( 5 )}

8200 = 6400 \cdot (1.054)^{4 x}

e^{4 x} - 7 e^{2 x} + 12 = 0

2^{- 3 d + 1} = 2^{- 2 d}

8^{4 x + 1} = 4^{3 x + 5}

5^{2 x + 5} = \frac{1}{5 ^{3 - x}}

What is the answer to x5^{2x+3}= 1/25 ?
The answer to x5^{2x+3}= 1/25 is x=(\W_{0}(\frac{2ln(5))/(3125))}{2ln(5)}
Solve for x: x5^{2x+3}= 1/25
The solution is x=(\W_{0}(\frac{2ln(5))/(3125))}{2ln(5)}