What is the answer to solvefor w,q=a*ln(bw+1) ?

The answer to solvefor w,q=a*ln(bw+1) is w=(e^{q/a}-1)/b

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solvefor w,q=a*ln(bw+1)

solve for

w, q = a \cdot ln (b w + 1)

w = \frac{e ^{\frac{q}{a}} - 1}{b}

Solution steps

q = a ln (b w + 1)

Switch sides

a ln (b w + 1) = q

Divide both sides by

a

ln (b w + 1) = \frac{q}{a}

Apply log rules

b w + 1 = e^{\frac{q}{a}}

Solve

b w + 1 = e^{\frac{q}{a}} : w = \frac{e ^{\frac{q}{a}} - 1}{b}

w = \frac{e ^{\frac{q}{a}} - 1}{b}

ln (Y) = 0

y, - ln (y + 6) = x^{2}

2 = lo g_{10} (2^{8} \frac{1}{x})

6 (ln (2 x) + 1) = 0

lo g_{x} (8) = \frac{3}{2}

What is the answer to solvefor w,q=a*ln(bw+1) ?
The answer to solvefor w,q=a*ln(bw+1) is w=(e^{q/a}-1)/b