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Popular >

solvefor y,x^y+z=0

Solution

solve for

y, x^{y} + z = 0

Solution

y = \frac{ln ( - z )}{ln ( x )}

Solution steps

x^{y} + z = 0

Subtract

z

from both sides

x^{y} + z - z = 0 - z

Simplify

x^{y} = - z

Apply exponent rules

y ln (x) = ln (- z)

Solve

y ln (x) = ln (- z) : y = \frac{ln ( - z )}{ln ( x )}

y = \frac{ln ( - z )}{ln ( x )}

Popular Examples

solvefor x,4^{x-1}=7^{2x}solve for

x, 4^{x - 1} = 7^{2 x}

9^2e^{2x}=81e^{2x}

9^{2} e^{2 x} = 81 e^{2 x}

(\frac{1}{5})^{2 x} = 125

3^{x+10}=9^{x+7}

3^{x + 10} = 9^{x + 7}

6(4^x)=5(3^{(2x-3)})

6 (4^{x}) = 5 (3^{(2 x - 3)})

Frequently Asked Questions (FAQ)

What is the answer to solvefor y,x^y+z=0 ?
The answer to solvefor y,x^y+z=0 is y=(ln(-z))/(ln(x))