What is the inverse of h(x)=log_{10}(x+1)-3 ?

The inverse of h(x)=log_{10}(x+1)-3 is 10^{x+3}-1

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inverse of h(x)=log_{10}(x+1)-3

inverse

h (x) = lo g_{10} (x + 1) - 3

1 0^{x + 3} - 1

Solution steps

y = lo g_{10} (x + 1) - 3

Replace

x

with

y

x = lo g_{10} (y + 1) - 3

Solve for

y, x = lo g_{10} (y + 1) - 3

1 0^{x + 3} - 1

+ 5 x + 4

f (t) = 300 \cdot 4^{t}

x + 7, x \geq - 7

g (x) = \frac{2 x + 1}{3 x + 4}

lo g_{a} (x)

What is the inverse of h(x)=log_{10}(x+1)-3 ?
The inverse of h(x)=log_{10}(x+1)-3 is 10^{x+3}-1