Math 1090
Quiz 5 - Extra credit
March 30, 2015
Answer the following questions in the space provided. Due: April 1.
Name:
UID:
1. Solve the following equations.
(a) (5 points) x2 ex − 5xex = 0
Solution. We start by factoring
x2 ex − 5xex = 0
xex (x − 5) = 0
For a product of three factors to be zero, one of them must necessarily be 0.

 x=0
ex = 0 =⇒ This can’t be the case, ex is never zero

x − 5 = 0 =⇒ x = 5
We conclude that the only possible solutions are x = 0 and x = 5.
(b) (5 points) log(x3 ) = (log x)3
Solution. By one of the properties of logarithms, we have that log(x3 ) = 3 log x.
Be careful: this property doesn’t apply to the right hand side.
log(x3 ) = (log x)3
3 log(x) = (log x)3
3 log(x) − (log x)3 = 0
log(x) 3 − (log x)2 = 0
As above, for a product of two factors to be zero, one of them must be zero, so we
have two possibilities:
log(x) = 0 =⇒ x = 100 = 1
√
√
3 − (log x)2 = 0 =⇒ (log x)2 = 3 =⇒ log x = ± 3 =⇒ x = 10± 3
√
√
We have thus found 3 possible solutions x = 1, 10 3 , 10− 3 . Remember that when
dealing with logarithmic equations, we must make sure that these solutions
make sense. We need to plug these solutions into our original equation and make
sure that they satisfy it. In this case, all three candidate solutions are actual
solutions.
(c) (5 points) ln(x2 − 4) + log3 (x2 − 4) = 1
Hint: Recall the base change formula: logb x =
loga x
.
loga b
Solution. The difficulty here resides in the fact the our equation contains two
logarithms written in different bases. We can remedy this by writing log3 (x2 − 4)
in base e, using the base change formula given in the hint, namely log3 (x2 − 4) =
ln(x2 −4)
.
ln 3
ln(x2 − 4) + log3 (x2 − 4) = 1
ln(x2 − 4)
ln(x2 − 4) +
=1
ln 3 1
ln(x2 − 4) 1 +
=1
ln 3
ln(x2 − 4) =
1
1 + ln13
ln(x2 − 4) = 0.5235
x2 − 4 = e0.5235
x2 − 4 = 1.6879
x2 = 1.6879 + 4 = 5.6879
√
x = ± 5.6879 = ±2.3849
We have found two candidate solutions x = ±2.3849. We need to plug them into
our original equation and make sure that everything makes sense (remember that
neither ln(x) nor log3 (x) are defined for negative values of x). In this case, both of
our solutions satisfy the original equation, indeed
ln(2.38492 − 4) + log3 (2.38492 − 4) = 0.9998 ' 1
ln (−2.3849)2 − 4 + log3 (−2.3849)2 − 4 = 0.9998 ' 1
Note that we are not getting exactly 1 because we are working with approximations
of the logarithm (we are working with 4 decimal digits only).
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