Illustrative
Mathematics
F-BF Exponentials and
Logarithms II
Alignments to Content Standards: F-BF.B.4.b
Task
Let f be the function defined by
g(x) = log10 (x).
f (x) = 10x and g be the function defined by
a. Sketch the graph of y
= f (g(x)). Explain your reasoning.
b. Sketch the graph of y
= g(f (x)). Explain your reasoning.
c. Let f and g be any two inverse functions. For which values of
which values of x does g(f (x)) = x?
x does f (g(x)) = x? For
IM Commentary
The purpose of this task is to help students see that the definition of the logarithm as
an exponent results in an inverse relationship between the exponential and logarithm
functions with the same base. This prepares them to think about the shape of the
graph of the logarithm function in terms of the shape of the graph of the exponential
function. As preparation for the reasoning required in this task, students might work
through Exponentials and Logarithms I.
Edit this solution
Solution
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Illustrative
Mathematics
a. In this case, we are trying sketch
f (g(x)) = 10log x
where log denotes the logarithm with base 10. We know that 10log x = x because log x
is, by definition, the exponent which 10 must be raised to in order to give a result of x.
Thus we must sketch a graph of f (g(x)) = x. The domain of g(x) = log x is all positive
real numbers and this is also the domain of f (g(x))
be a graph of y = x with domain x > 0:
= 10log x . Thus our graph should
b. Here we are trying to sketch
g(f (x)) = log 10x
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Illustrative
Mathematics
where the logarithm is with base 10. By definition, log 10x is the exponent
raised to in order to give a result of 10x so
10 must be
log 10x = x.
There is no restraint on the domain of g(f (x)) = log 10x because the domain of
f (x) = 10x is any real number and its range is all positive real numbers. So all the
values of f (x) are in the domain of g(x) = logb (x). Thus our graph is a graph of y =
with no restriction on the domain:
x
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Illustrative
Mathematics
c. Generalizing what we saw in parts (a) and (b), if f and g are inverse functions then
g(f (x)) = x for all values of x in the domain of f and
f (g(x)) = x for all values of x in the domain of g.
For example, in part (a), we see that f (g(−1))
of g.
≠ −1, because −1 is not in the domain
F-BF Exponentials and Logarithms II
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