What is the answer to solvefor k,log_{k}(a)=b-2 ?

The answer to solvefor k,log_{k}(a)=b-2 is k=e^{(ln(a))/(b-2)}

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solvefor k,log_{k}(a)=b-2

solve for

k, lo g_{k} (a) = b - 2

k = e^{\frac{l n ( a )}{b - 2}}

Solution steps

lo g_{k} (a) = b - 2

Apply log rules

\frac{ln ( a )}{ln ( k )} = b - 2

Multiply both sides by

ln (k)

\frac{ln ( a )}{ln ( k )} ln (k) = b ln (k) - 2 ln (k)

Simplify

ln (a) = b ln (k) - 2 ln (k)

Switch sides

b ln (k) - 2 ln (k) = ln (a)

Factor

b ln (k) - 2 ln (k) : ln (k) (b - 2)

ln (k) (b - 2) = ln (a)

Divide both sides by

b - 2

ln (k) = \frac{ln ( a )}{b - 2}

Apply log rules

k = e^{\frac{l n ( a )}{b - 2}}

lo g_{x} (3) = 31

lo g_{x} (8) = 1 \cdot 5

y, ln (x^{3} y^{3}) = c

x + ln (x) = - 4

6 ln (x + 2) - 3 = 15

What is the answer to solvefor k,log_{k}(a)=b-2 ?
The answer to solvefor k,log_{k}(a)=b-2 is k=e^{(ln(a))/(b-2)}