What is the answer to 5^x-24=(5^2)/5 x ?

The answer to 5^x-24=(5^2)/5 x is

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5^x-24=(5^2)/5 x

5^{x} - 24 = \frac{5 ^{2}}{5} x

x = - \frac{5 W _{- 1} ( - \frac{l n ( 5 )}{5 e ^{\frac{24 l n ( 5 )}{5}}} ) + 24 ln ( 5 )}{5 ln ( 5 )}, x = - \frac{5 W _{0} ( - \frac{l n ( 5 )}{5 e ^{\frac{24 l n ( 5 )}{5}}} ) + 24 ln ( 5 )}{5 ln ( 5 )}

Decimal Notation

x = 2.20994 \dots, x = - 4.79991 \dots

Solution steps

5^{x} - 24 = \frac{5 ^{2}}{5} x

Prepare

5^{x} - 24 = \frac{5 ^{2}}{5} x

for Lambert form:

1 = (5 x + 24) e^{- l n (5) x}

Rewrite the equation with

(- x - \frac{24}{5}) ln (5) = u

and

x = - \frac{5 u + 24 ln ( 5 )}{5 ln ( 5 )}

1 = (5 (- \frac{5 u + 24 ln ( 5 )}{5 ln ( 5 )}) + 24) e^{- l n (5) (- \frac{5 u + 24 l n ( 5 )}{5 l n ( 5 )})}

Simplify

1 = e^{\frac{5 u + 24 l n ( 5 )}{5}} (- \frac{5 u + 24 ln ( 5 )}{ln ( 5 )} + 24)

Rewrite

1 = e^{\frac{5 u + 24 l n ( 5 )}{5}} (- \frac{5 u + 24 ln ( 5 )}{ln ( 5 )} + 24)

in Lambert form:

\frac{e ^{\frac{5 u + 24 l n ( 5 )}{5}} u}{e ^{\frac{24 l n ( 5 )}{5}}} = - \frac{ln ( 5 )}{5 e ^{\frac{24 l n ( 5 )}{5}}}

Solve

\frac{e ^{\frac{5 u + 24 l n ( 5 )}{5}} u}{e ^{\frac{24 l n ( 5 )}{5}}} = - \frac{ln ( 5 )}{5 e ^{\frac{24 l n ( 5 )}{5}}} : u = W_{- 1} (- \frac{ln ( 5 )}{5 e ^{\frac{24 l n ( 5 )}{5}}}), u = W_{0} (- \frac{ln ( 5 )}{5 e ^{\frac{24 l n ( 5 )}{5}}})

u = W_{- 1} (- \frac{ln ( 5 )}{5 e ^{\frac{24 l n ( 5 )}{5}}}), u = W_{0} (- \frac{ln ( 5 )}{5 e ^{\frac{24 l n ( 5 )}{5}}})

Substitute back

u = (- x - \frac{24}{5}) ln (5),

solve for

x

Solve

(- x - \frac{24}{5}) ln (5) = W_{- 1} (- \frac{ln ( 5 )}{5 e ^{\frac{24 l n ( 5 )}{5}}}) : x = - \frac{5 W _{- 1} ( - \frac{l n ( 5 )}{5 e ^{\frac{24 l n ( 5 )}{5}}} ) + 24 ln ( 5 )}{5 ln ( 5 )}

Solve

(- x - \frac{24}{5}) ln (5) = W_{0} (- \frac{ln ( 5 )}{5 e ^{\frac{24 l n ( 5 )}{5}}}) : x = - \frac{5 W _{0} ( - \frac{l n ( 5 )}{5 e ^{\frac{24 l n ( 5 )}{5}}} ) + 24 ln ( 5 )}{5 ln ( 5 )}

x = - \frac{5 W _{- 1} ( - \frac{l n ( 5 )}{5 e ^{\frac{24 l n ( 5 )}{5}}} ) + 24 ln ( 5 )}{5 ln ( 5 )}, x = - \frac{5 W _{0} ( - \frac{l n ( 5 )}{5 e ^{\frac{24 l n ( 5 )}{5}}} ) + 24 ln ( 5 )}{5 ln ( 5 )}

5^{(x - 1)} + 5^{(x - 2)} = 30

(\frac{1}{2})^{\frac{t}{2}} = 64

3^{2 x - 1} \cdot 2 7^{3 - x} = 1

3 x + 2 = e^{x}

4^{x} = \frac{1}{8}