What is the inverse of log_{10}(2)(kx-10) ?

The inverse of log_{10}(2)(kx-10) is (x+10log_{10}(2))/(log_{10)(2)k}

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inverse of log_{10}(2)(kx-10)

inverse

lo g_{10} (2) (k x - 10)

\frac{x + 10 lo g _{10} ( 2 )}{lo g _{10} ( 2 ) k}

Solution steps

y = lo g_{10} (2) (k x - 10)

Replace

x

with

y

x = lo g_{10} (2) (k y - 10)

Solve for

y, x = lo g_{10} (2) (k y - 10)

\frac{x + 10 lo g _{10} ( 2 )}{lo g _{10} ( 2 ) k}

f (x) = 4 x^{\frac{1}{5}} + 2

f (x) = 121 x^{2}

f (x) = 13 - x^{2}

f (x) = \frac{x + 3}{7}

f (x) = \frac{1}{3} (5)^{- 2 (x + 4)} - 1

What is the inverse of log_{10}(2)(kx-10) ?
The inverse of log_{10}(2)(kx-10) is (x+10log_{10}(2))/(log_{10)(2)k}