What is the answer to ln(15)=ln(-n)-nln(1.1) ?

The answer to ln(15)=ln(-n)-nln(1.1) is n=-(\W_{0}(15ln(1.1)))/(ln(1.1))

Solve for n: ln(15)=ln(-n)-nln(1.1)

The solution is n=-(\W_{0}(15ln(1.1)))/(ln(1.1))

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ln (15) = ln (- n) - n ln (1.1)

n = - \frac{W _{0} ( 15 ln ( 1.1 ) )}{ln ( 1.1 )}

Decimal Notation

n = - 7.40553 \dots

Solution steps

ln (15) = ln (- n) - n ln (1.1)

Switch sides

ln (- n) - n ln (1.1) = ln (15)

Move

n ln (1.1)

to the right side

ln (- n) = ln (15) + n ln (1.1)

f (x) = g (x),

then

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

e^{l n (- n)} = e^{l n (15) + n l n (1.1)}

Simplify

e^{l n (- n)} : - n

Simplify

e^{l n (15) + n l n (1.1)} : 15 \cdot 1. 1^{n}

- n = 15 \cdot 1. 1^{n}

Solve

- n = 15 \cdot 1. 1^{n} : n = - \frac{W _{0} ( 15 ln ( 1.1 ) )}{ln ( 1.1 )}

n = - \frac{W _{0} ( 15 ln ( 1.1 ) )}{ln ( 1.1 )}

Verify Solutions:

n = - \frac{W _{0} ( 15 ln ( 1.1 ) )}{ln ( 1.1 )}

True

The solution is

n = - \frac{W _{0} ( 15 ln ( 1.1 ) )}{ln ( 1.1 )}

lo g_{2} (x + 7) + lo g_{2} (x - 7) = 5

5 ln (3 x - 4) = 30

3^{l o g_{3} (3 x + 4)} = 3 x

15 = 10 \cdot lo g_{10} (x)

ln (3 x - 18) = 4

What is the answer to ln(15)=ln(-n)-nln(1.1) ?
The answer to ln(15)=ln(-n)-nln(1.1) is n=-(\W_{0}(15ln(1.1)))/(ln(1.1))
Solve for n: ln(15)=ln(-n)-nln(1.1)
The solution is n=-(\W_{0}(15ln(1.1)))/(ln(1.1))