What is the answer to ln(250r)+ln(1)=42r ?

The answer to ln(250r)+ln(1)=42r is r=-(\W_{-1}(-21/125))/(42),r=-(\W_{0}(-21/125))/(42)

Solve for r: ln(250r)+ln(1)=42r

The solution is r=-(\W_{-1}(-21/125))/(42),r=-(\W_{0}(-21/125))/(42)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

ln (250 r) + ln (1) = 42 r

r = - \frac{W _{- 1} ( - \frac{21}{125} )}{42}, r = - \frac{W _{0} ( - \frac{21}{125} )}{42}

Decimal Notation

r = 0.06716 \dots, r = 0.00491 \dots

Solution steps

ln (250 r) + ln (1) = 42 r

Move

ln (1)

to the right side

ln (250 r) = 42 r - ln (1)

Apply log rules

ln (250 r) = 42 r

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

e^{l n (250 r)} = e^{42 r}

Simplify

e^{l n (250 r)} : 250 r

250 r = e^{42 r}

Solve

250 r = e^{42 r} : r = - \frac{W _{- 1} ( - \frac{21}{125} )}{42}, r = - \frac{W _{0} ( - \frac{21}{125} )}{42}

r = - \frac{W _{- 1} ( - \frac{21}{125} )}{42}, r = - \frac{W _{0} ( - \frac{21}{125} )}{42}

Verify Solutions:

r = - \frac{W _{- 1} ( - \frac{21}{125} )}{42}

True

, r = - \frac{W _{0} ( - \frac{21}{125} )}{42}

True

The solutions are

r = - \frac{W _{- 1} ( - \frac{21}{125} )}{42}, r = - \frac{W _{0} ( - \frac{21}{125} )}{42}

lo g_{12} (5 b - 8) = lo g_{12} (3 b)

lo g_{5} ((x + 1)^{4}) = 12

lo g_{2} (2 x - 5) + lo g_{2} (x + 1) = 1

- lo g_{10} (x) = 5.59

3 ln (4 x) = - 3

What is the answer to ln(250r)+ln(1)=42r ?
The answer to ln(250r)+ln(1)=42r is r=-(\W_{-1}(-21/125))/(42),r=-(\W_{0}(-21/125))/(42)
Solve for r: ln(250r)+ln(1)=42r
The solution is r=-(\W_{-1}(-21/125))/(42),r=-(\W_{0}(-21/125))/(42)