What is the inverse of h(x)=log_{10}(2x+3) ?

The inverse of h(x)=log_{10}(2x+3) is (10^x-3)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of h(x)=log_{10}(2x+3)

inverse

h (x) = lo g_{10} (2 x + 3)

\frac{1 0 ^{x} - 3}{2}

Solution steps

y = lo g_{10} (2 x + 3)

Replace

x

with

y

x = lo g_{10} (2 y + 3)

Solve for

y, x = lo g_{10} (2 y + 3)

\frac{1 0 ^{x} - 3}{2}

f (x) = (x + 4)^{3} + 8

\frac{21}{x ^{2}}

\frac{1}{8} x^{3} - 3

\frac{x ^{2} + 13 x + 42}{x + 7}

f (x) = \frac{5 x + 3}{3 x - 7}

What is the inverse of h(x)=log_{10}(2x+3) ?
The inverse of h(x)=log_{10}(2x+3) is (10^x-3)/2