Compositions of Logarithmic and Exponential Functions
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Compositions of Logarithmic and Exponential Functions

August 24, 2019


>>Here we’re going to find
the composition of f with g and the composition of g
with f, and their domains, where f of x is, 2 to the
x and g of x is, x plus 4. So first of all, we’ll
look at f of g of x. Which is what f takes g of x to. Since g of x is, x plus 4 and f
uses its input as the exponent on 2, we get 2 to
the power, x plus 4. There is no restriction
when x goes to, x plus 4, or when we exponentiate.
Both, have domains which are all reals, and
so the domain of the composition of f with g is equal to
negative infinity, to infinity. Now as we look at g..
composition of g with f, of x. That is what
g takes f of x to. Where f of x is, 2 to the x. And
what does g to do to anything? g adds 4, and so
we’re going to have 2 to the x quantity, plus 4. This time x first of
all gets exponentiated, and.. I’m sorry.. I wrote
e instead of 2 here. And then we add 4, and so
there’s no restriction on either of these, the domain
of the composition of g with f is negative
infinity, to infinity. In our second problem, f
of x is log base 4 of x, and g of x is, x minus 4. So the composition of f with g,
of x, is what f takes g of x to.. what f takes, x minus 4 to and f takes the log base
4 of whatever it sees. So we get a log base
4 of, x minus 4. So x first of all
goes to, x minus 4, which allows any real number. But then to log base
4 of, x minus 4. And here we have a restriction.
x minus 4, must be greater than 0 because the log only
accepts positive values. So x minus 4, greater than
0 implies x must be greater than 4, and so the domain of the
composition of f with g is equal to the open interval 4 to
infinity, not including the 4. For the composition
of g with f of x, the result is what
g takes f of x to. And so this will be g of log
base 4 of x, g subtracts 4 from anything it sees. So it’s the log base 4 of
x quantity, subtract 4. This time x first of all is
taken to log base 4 of x, and then 4 is subtracted
from log base 4 of x. So the restrictions
occurs in the first step, where x must be positive. Because the domain of the
log base 4 is 0 to infinity, and so the domain of
the composition of g with f is the open
interval from 0 to infinity. There’s no further restriction because we can subtract
4 from any real number. And our last problem, we have f
of x equals log x.. common log x.. and g of x equals
x to the fifth.. it’s called a power
function.. x to the fifth, where the variable
is in the base. So composition of f
with g of x takes x to what f takes g of x to. And g of x is x to the fifth. f is the common log, so we end up with the log of
x to the fifth. So x goes to x to the fifth, which goes to the log
of x to the fifth. So x to x to the fifth, x
could be any real number, but in the second step x to the
fifth must be greater than 0. If x to the fifth
is greater than 0, we notice that a negative number
to the fifth would be negative, and so we must have
x greater than 0, because we’re not allowed to have a negative power
as an input to log. And so the domain of
the composition of f with g is the open
interval from 0 to infinity. In the other direction we have
the composition of g with f of x, which is what
g takes f of x to, and f of x is the
common log of x. And g raises whatever it
sees to the fifth power. So we’re going to have
the common log of x, parentheses to the fifth power. So x goes to log x, which goes to log x quantity
to the fifth power. Here the restriction occurs
right away, because the input to the log has to
be greater than 0. We can raise any number
to the fifth power, so there’s no further
restriction, and so the domain of the composition of g with f is the open
interval from 0 to infinity.

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