What is the answer to 5^x-5^5-x=150 ?

The answer to 5^x-5^5-x=150 is x=-(\W(-\frac{ln(5))/(5^{3275)})+3275ln(5)}{ln(5)}

Solve for x: 5^x-5^5-x=150

The solution is x=-(\W(-\frac{ln(5))/(5^{3275)})+3275ln(5)}{ln(5)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5^{x} - 5^{5} - x = 150

x = - \frac{W ( - \frac{l n ( 5 )}{5 ^{3275}} ) + 3275 ln ( 5 )}{ln ( 5 )}

Solution steps

5^{x} - 5^{5} - x = 150

Prepare

5^{x} - 5^{5} - x = 150

for Lambert form:

1 = (x + 3275) e^{- l n (5) x}

Rewrite the equation with

(- x - 3275) ln (5) = u

and

x = - \frac{u + 3275 ln ( 5 )}{ln ( 5 )}

1 = ((- \frac{u + 3275 ln ( 5 )}{ln ( 5 )}) + 3275) e^{- l n (5) (- \frac{u + 3275 l n ( 5 )}{l n ( 5 )})}

Simplify

1 = e^{u + 3275 l n (5)} (- \frac{u + 3275 ln ( 5 )}{ln ( 5 )} + 3275)

Rewrite

1 = e^{u + 3275 l n (5)} (- \frac{u + 3275 ln ( 5 )}{ln ( 5 )} + 3275)

in Lambert form:

e^{u} u = - \frac{ln ( 5 )}{5 ^{3275}}

Solve

e^{u} u = - \frac{ln ( 5 )}{5 ^{3275}} : u = W (- \frac{ln ( 5 )}{5 ^{3275}})

u = W (- \frac{ln ( 5 )}{5 ^{3275}})

Substitute back

u = (- x - 3275) ln (5),

solve for

x

Solve

(- x - 3275) ln (5) = W (- \frac{ln ( 5 )}{5 ^{3275}}) : x = - \frac{W ( - \frac{l n ( 5 )}{5 ^{3275}} ) + 3275 ln ( 5 )}{ln ( 5 )}

x = - \frac{W ( - \frac{l n ( 5 )}{5 ^{3275}} ) + 3275 ln ( 5 )}{ln ( 5 )}

x, 2^{x} \cdot 3^{y} = 12

6^{n + 7} = 67

1 2^{5 x} = 3 \cdot 1 5^{3 x}

3^{x - 3} = 1 7^{- 5 x}

9 e^{x} = 3

What is the answer to 5^x-5^5-x=150 ?
The answer to 5^x-5^5-x=150 is x=-(\W(-\frac{ln(5))/(5^{3275)})+3275ln(5)}{ln(5)}
Solve for x: 5^x-5^5-x=150
The solution is x=-(\W(-\frac{ln(5))/(5^{3275)})+3275ln(5)}{ln(5)}