- d.The graph off(x) = –e2–x– 3 can be
obtained by reflecting the graph ofg(x) = e–x
about the y-axis, then shifting it to the left 2
units, then reflecting it over the x-axis, and
finally shifting it down 3 units. The graph is as
follows:
It is evident that all three characteristics pro-
vided in choices,a,b, and c hold.
- d.Since 8 can be expressed as a power of 2, we
can write the following:
2 x–5 = 8
2 x–5 = 2^3
Equating exponents yields x– 5 = 3, so x= 8 is
the solution.
- b.First, rewrite 9 as a power of 3. Then, sim-
plify the right side, as follows:
32 x= 9 3 x–1
32 x= 3^2 3 x–1
32 x= 32+x–1
32 x= 3x+1
Equating exponents yields 2x= x+ 1. Solving
this equation yields x= 1.
- c.Multiply both sides by the denominator on
the right side. Then, observe that 1 = 4^0.
42 x–3 = 41 x
42 x–3 4 x = 41 x 4 x
42 x–3+x= 1
43 x–3 = 4^0
3 x– 3 = 0
x= 1
- d.The power to which 125 should be raised in
order to yield 25 is not obvious. However, note
that 125 and 25 are both powers of 5. We use
this fact in order to first rewrite the equation in
a more convenient form, and then solve for x:
125 x= 25
(5^3 )x= 5^2
53 x= 5^2
Equating exponents yields 3x= 2. Thus, we
conclude that the solution is x= ^23 .
- b.First, simplify the left side. Then, equate the
exponents and solve the resulting equation.
(ex)x–3= e^10
ex2–3x= e^10
x^2 – 3x= 10
x^2 – 3x– 10 = 0
(x– 5)(x+ 2) = 0
x= 5,x= –2
- d.We rewrite the left side of the equation as
a power of 4, then equate the exponents and
solve for x:
163 x–1= 4^2 x+3
(4^2 )^3 x–1= 4^2 x+3
4 2(3x–1)= 4^2 x+3
2(3x– 1) = 2x+ 3
6 x– 2 = 2x+ 3
4 x= 5
x= ^54
–6 –4 4 6
–4
–2
–2
–6
–8
–10
–12
–14
4
2
2
ANSWERS & EXPLANATIONS–
