f(x) = b^x and g(x) = log b X are inverse functions. Explain why each of the following are true.

1. A translation of function f is f1(x) = b^(x-h). It is equivalent to a vertical stretch or vertical compression of function f.

2. The inverse of f1(x) = b^(x-h) is not equivalent to a translation of g.

3. The inverse of f1 (x) =b^(x-h) is not equivalent to a vertical stretch or vertical compression of g.

4. The function h(x) = log c X is a vertical stretch or compression of g or of its reflection -g. Read this as"negative g"

Will probably needs to use the properties of exponents and logarithms and change of base formulas to change the functions into alternate forms

If we used the rules of powers, we know that when we have a subtraction in the power of an exponential, it can be split into a division. So the function can be rewritten as:

[tex]f_{1}(x)=\frac{b^{x}}{b^{h}}[/tex]

Remember the original function was:

[tex] f(x)=b^{x}[/tex]

therefore:

[tex]f_{1}(x)=\frac{f(x)}{b^{h}}[/tex]

this means that if [tex]b^{h}<1[/tex] then it will be a vertical stretch.

If [tex]b^{h}>1[/tex] then it will be a vertical stretch.

2. The inverse of [tex]f_{1}(x)=b^{x-h}[/tex] is not equivalent to a translation of g.

This is partially true and you'll see why. Let's start by finding the inverse of that function:

[tex]x=b^{y-h}[/tex]

we start by turning the given power to a multiplication of powers so we get:

[tex]x=b^{-h}b^{y}[/tex]

we then move the [tex]b^{-h}[/tex] to the other side of the equation so we get:

[tex]xb^{h}=b^{y}[/tex]

and turn the equation into a logarithm:

[tex]y=log_{b}(xb^{h})[/tex]

so:

[tex]g_{1}(x)=log_{b}(xb^{h})[/tex]

or:

[tex]g_{1}(x)=g(b^{h}x)[/tex]

remember that when you multiply a constant by x, you will get a horizontal compression if [tex]b^{h}>1[/tex] and a horizontal stretch if [tex]b^{h}<1[/tex].

but there is another interpretation for this function. Let's take the original equation:

[tex]x=b^{y-h}[/tex]

if we directly turned this equation into a logarithm we would get that:

[tex]y-h=log_{b}(x)[/tex]

so:

[tex]y=h+log_{b}(x)[/tex]

or:

[tex]g_{1}(x)=h+g(x)[/tex]

if the inverse function is written like this, it can be interpreted as a vertical shift. Both interpretations are correct.

3. The inverse of [tex]f_{1}(x)=b^{x-h}[/tex] is not equivalent to a vertical stretch or vertical compression of g.

As we saw in the previous part of the problem, that function is either a horizontal stretch/compression or a vertical shift, not a vertical stretch or compression.

4. The function [tex]h(x)=log_{c}(x)[/tex] is a vertical stretch or compression of g or of its reflection -g.

We can rewrite the function like this thanks to log rules:

[tex]h(x)=\frac{log_{b}(x)}{log_{b}(c)}[/tex]

which is the same as:

[tex]h(x)=\frac{g(x)}{log_{b}(c)}[/tex]

If [tex]log_{b}(c)>1[/tex] it will be a vertical compression of g(x). If [tex]log_{b}(c)<1[/tex], then it will be a vertical stretch no matter if g is positive or negative.