Worksheet 3.5A, Solving Log Equations
MATH 1410
(SOLUTIONS)
1. Solve for x in the following equations, finding the exact value of x. Then use your calculator to
approximate the value of x to four decimal places.
(a) 2x = 17
(b) 2x = 3x+1
(c) 5x+1 = 32x−2
(d) 43x = 22x+2
Solution.
(a) (See class notes.) x = log2 (17) =
ln 17
≈ 4.08746284 .
ln 2
(b) (See class notes.) Take the natural log of both sides of the equation 2x = 3x+1 to obtain
ln(2x ) = ln(3x+1 )
Rewrite this as
x ln 2 = (x + 1) ln 3
so
x ln 2 = x ln 3 + ln 3.
Move terms with an x to the leftside:
x ln 2 − x ln 3 = ln 3
and factor out x.
x(ln 2 − ln 3) = ln 3.
Finally, solve for x:
x=
ln 3
≈ −2.70951129 .
ln 2 − ln 3
(c) To solve 5x+1 = 32x−2 , take the natural log of both sides and use the exponent rule to move
the exponent down so that
(x + 1) ln 5 = (2x − 2) ln 3
Collecting terms with an x on the left-hand side of the equation we have
x ln 5 − 2x ln 3 = − ln 5 − 2 ln 3.
Factoring out the x on the left-hand side:
x(ln 5 − 2 ln 3) = − ln 5 − 2 ln 3
and so
x=
− ln 5 − 2 ln 3
ln 5 + 2 ln 3
=
≈ 6.47626548
ln 5 − 2 ln 3
2 ln 3 − ln 5
(d) 43x = 22x+2 . We can avoid logarithms completely in this problem if we are willing to write
the two exponential forms in the same base. Since 4 = 22 then we can rewrite our equation as
(22 )3x = 22x+2
and so
26x = 22x+2 .
Since the bases are the same then the exponents must be the same and so
6x = 2x + 2
This implies
1
.
2
4x = 2 =⇒ x =
(Let’s check our work here: If x =
1
then the left-hand side of the original equation is
2
3
1
43x = 4 2 = (4 2 )3 = 23 = 8.
On the other hand, the right-hand side is then
1
2(2· 2 +2) = 23 = 8.
So the two sides agree and so yes, our solution is x =
1
.)
2
2. Solve for x in the following equations, finding the exact value of x. Then use your calculator to
approximate the value of x to four decimal places.
(a) log x + log(x − 9) = 1.
(b) log3 x + log3 (x − 6) = 1.
(c) log3 (2x2 − 8) − log3 (x − 2) = 4.
Solution.
(a) log x + log(x − 9) = 1 implies, by the product property, that log(x(x − 9)) = 1. Put this
expression into exponential form we have that
x(x − 9) = 101 .
So
x2 − 9x = 10.
Getting a zero on the right-hand side we have
x2 − 9x − 10 = 0.
The left-hand side factors as (x − 10)(x + 1) so
(x − 10)(x + 1) = 0
and so either x = 10 or x = −1. We should check these two answers in the original equation.
If we do that, we see that log −1 does not make sense and so x cannot be −1. So our final
answer is x = 10 .
(b) log3 x + log3 (x − 6) = 1 implies, by the product property, that log3 x(x − 6) = 1. Putting this
equation into exponential form gives us
x(x − 6) = 31
so
x2 − 6x = 3
so
x2 − 6x − 3 = 0.
By the quadratic formula we have
√
√
√
√
6 ± 48
6±4 3
6 ± 36 + 12
=
=
= 3 ± 2 3.
x=
2
2
2
√
√
So our two answers are x = 3 + 2 3 ≈ 6.4641 and 3 ± 2 3 ≈ −0.4641. If we check these two
answers in the original equation we discover that the negative solution won’t work and so the
√
answer must be 3 + 2 3 ≈ 6.4641 .
(c) (See class notes.) We solve log3 (2x2 − 8) − log3 (x − 2) = 4 by using our “division property”
2x2 − 8
to rewrite the lefthand side as log3
. We may factor 2x2 − 8 as 2(x − 2)(x + 2) and so
x−2
2x2 − 8
(as long as x 6= 2) simplify
= 2(x + 2) = 2x + 4. So out equation simplifies to
x−2
log3 (2x + 4) = 4.
Putting this log equation into exponential form, one can write
2x + 4 = 34 = 81.
We can (easily!) solve equations like 2x + 4 = 81 and get x =
77
.
2
3. Use logarithms to find the number of digits in the following large prime numbers:
(a) 237156667 − 1
(b) 232582657 − 1
(c) 230402457 − 1
(d) 225964951 − 1
(e) 224036583 − 1
(f) 220996011 − 1
(g) 213466917 − 1
Solutions.
(a) log(237156667 − 1) ≈ log(237156667 ) = 37156667 log(2) = 37156667(
the number of digits should be 11185272 , over 11 million digits!
(b) 232582657 − 1 has 9808358
(c) 2
30402457
− 1 has 9152052
digits.
digits.
ln 10
) ≈ 11185271.306. So
ln 2
(d) 225964951 − 1 has 7816230
digits.
(e) 224036583 − 1 has 7235733
digits.
20996011
− 1 has 6320430
digits.
(g) 213466917 − 1 has 4053946
digits.
(f) 2
4. $1000 is invested at 4% annual interest compounded continuously.
(a) What is the future value of this investment after seven years?
(b) When does this investment reach $1600?
(c) When does this investment double (that is, reach $2000)?
Solutions.
(a) 1000(e0.04·7 ) = 1000(e0.28 ) ≈ 1323.13 so the future value is $1323.13
(b) Solve for t in the equation 1600 = 1000(e0.04t ) Divide both sides by 1000 to obtain the equation
1.6 = (e0.04t )
and then take the natural log of both sides:
ln(1.6) = 0.04t.
So
ln(1.6)
≈ 11.7501 years ,
0.04
which is slightly more than 11 years and 9 months.
t=
(c) Solve for t in the equation 2000 = 1000(e0.04t ) Divide both sides by 1000 to obtain the equation
2 = (e0.04t )
and then take the natural log of both sides:
ln(2) = 0.04t.
So
ln(2)
≈ 17.3287 years ,
0.04
which is just under 17 years and 4 months.
t=
Comment.
Notice how compounding continuously generates money just slightly faster than compounding monthly.
The future value of the continuous compounding, after seven years, is about sixty cents more valuable
than compounding monthly. And the doubling times are almost the same, with compounding continously
being slightly quicker, by a few weeks across more than 17 years.