Section 6.6 solutions
#1 - 12: Solve the exponential equation by writing each side of the equation with the same base then
equating the exponents. Problems also may be solved with logarithms.
1) 2x = 16 (I will rewrite the 16 as 24)
2x = 24
Answer: x = 4
3) 2x+1 = 32 (rewrite the 32 as 25)
2x+1 = 25
x+1=5
Answer: x = 4
1 𝑥
5) (2) = 16 (easier to do with logs than my method)
Rewrite
1
2
𝑎𝑠 2−1 𝑎𝑛𝑑 16 𝑎𝑠 24
(2-1)x = 24
2-x = 24
-x = 4
Answer: x = -4
7) 24−𝑥 = 64 (rewrite 64 as 26)
24-x = 26
4–x=6
-x = 2
Answer: x = -2
9) 32x = 2 (rewrite 32 as 25)
(25)x = 21
25x = 21
5x = 1
Answer: x = 1/5
11) 16x = 4 (rewrite 16 as 42)
(42)x = 41
42x = 41
2x = 1
Answer: x = 1/2
#13 - 24: Solve the exponential equations, round your answer to 2 decimals.
13) log (3x )= log 6 (take log of each side)
x[log(3)] = log 6 ( use power to product rule)
𝑥=
log 6
log 3
(divide then use calculator)
Answer: x = 1.63
15) ex = 12 (take ln of each side)
Ln ex = ln 12
x(ln e)= ln 12 (use power to product rule)
x = ln 12 (ln e = 1)
Answer: x = ln(12) = 2.48
17) 5(10x) = 20 (divide both sides by 5)
10x = 4 (now take log of each side)
Log 10x = log 4 (use power to product rule)
x(log 10) = log 4 (log 10 = 1)
x = log 4
Answer: x = log(4) = .60
19) 32e2x = 128 (divide each side by 32)
e2x = 4 (take ln of each side)
ln e2x = ln 4 (use power to product rule)
2x(ln e)= ln 4 (ln e = 1)
2x = ln 4 (divide by 2)
Answer: x =
𝐥𝐧(𝟒)
𝟐
= .69
21) 3x-1 = 5x (take log of both sides)
Log(3x-1) = log(5x) (use power to product rule)
(x-1)log(3) = x(log 5)
x( log 3) – log(3) = x( log 5) (move the xlog(5) to left and –log(3) to right)
x(log 3) – x(log 5)= log(3) (factor out GCF of x)
x( log 3 – log 5 )= log(3) (divide)
Answer: 𝒙 =
𝐥𝐨𝐠(𝟑)
𝐥𝐨𝐠(𝟑)−𝐥𝐨𝐠(𝟓)
= −𝟐. 𝟏𝟓
23) 6x-1 = 9x (take log of both sides)
Log(6x-1) = log(9x)
(use power to product rule)
(x-1)log 6 = x(log 9 )
x(log 6) – log(6) = x(log 9) (move the x(log 9) to left and –log(6) to right)
x(log 6) – x(log 9) = log(6) (factor out GCF of x)
x( log(6) – log(9) ) = log(6) (divide)
Answer: 𝒙 =
𝐥𝐨𝐠(𝟔)
𝐥𝐨𝐠(𝟔)−𝐥𝐨𝐠(𝟗)
= −𝟒. 𝟒𝟐
#25 - 51: Solve the logarithmic equations, round to 2 decimals when needed.
25) log3 x = 2
32 = x
Answer: x = 9
27) ln x = 1 (write ln x as logex, then write in exponential form)
Log ex = 1
Answer: x = e1 = 2.72
29) logx 49 = 2 (x >0)
x2 = 49 (take the square root of both sides, don’t need ± since answer can only be positive)
Answer: x = 7
31) logx 64 = 3 (x >0)
x3 = 64 (now take cubed root of each side)
3
3
√𝑥 3 = √64
Answer: x = 4
33) logx 3 =
𝑥
1⁄
2
1
2
(x>0)
= 3 (write as square root since ½ power and square root are equivalent)
√𝑥 = 3
2
(√𝑥) = 32
Answer: x = 9
35) log2(x-1) = 3
23 = x-1
8=x–1
Answer: x = 9
37) log2(2x)=5
25 = 2x
32 = 2x
Answer: x = 16
39) log (x+1) = log (3x–2) (drop logs, set arguments equal)
x + 1 = 3x – 2
-2x = -3
Answer: x = 3/2
41) log2 (x+3) = log2 (3x) (drop logs, set arguments equal)
x + 3 = 3x
3 = 2x
Answer: x = 3/2
43) log 2x – log2(x+6)=-2 (use minus to divide rule to rewrite left side with one log)
𝑙𝑜𝑔2
𝑥
𝑥+6
= −2 (write in exponential form)
𝑥
2−2 = 𝑥+6 (write 2-2 as ½2)
1
4
𝑥
= 𝑥+6 (cross multiply)
4x = x + 6
3x = 6
Answer: x = 2
45) log2 x – log2 (x – 6) = 2 (use minus to divide rule to rewrite left side with one log)
𝑙𝑜𝑔2
𝑥
𝑥−6
= 2 (write in exponential form)
𝑥
22 = 𝑥−6 (rewrite the 22 as 4, then as a fraction over 1)
4
1
𝑥
= 𝑥−6 (cross multiply)
4x – 24 = x
3x = 24
Answer: x = 8
47) log2 (x+6) – log2 (3x+2)=-1 (use minus to divide rule to rewrite left side with one log)
𝑥+6
𝑙𝑜𝑔2 3𝑥+2 = −1
𝑥+6
2−1 = 3𝑥+2 (write 2-1 as ½1)
1
2
𝑥+6
= 3𝑥+2 (cross multiply)
3x + 2 = 2x + 12
Answer: x = 10
49) log3 x + log3 (x+6) = 3 (use the plus to multiply rule to rewrite left side with one log)
Log3 (x)(x+6) = 3
Log3(x2 + 6x) = 3 (write in exponential form)
33 = x2 + 6x
27 = x2 + 6x
0 = x2 + 6x – 27
0 = (x+9)(x-3)
x+9=0
x–3=0
x = -9
x=3
Check x = -9
Log3(-9) + log3(-9+6) =3
I get i on the left side, so this doesn’t check and -9 is not a solution
Check x = 3
Log3 3 + log3 (3+6) = 3
1 + 2 = 3 (3 is a solution)
Answer: x = 3
(no need to mention the x =-9 as it doesn’t check
It would be okay to write: Answer: x = 3, x = -9 is extraneous)
51) log3 (x+6) + log3 (3x) = 4 (use the plus to multiply rule to rewrite left side with one log)
Log3(x+6)(3x) = 4
Log3 (3x2 + 18x) = 4
34 = 3x2 + 18x
81 = 3x2 + 18x (divide every term by 3 to make numbers smaller)
27 = x2 + 6x
0 = x2 + 6x – 27
0 = (x+9)(x-3)
x+9=0
x–3=0
x = -9
x=3
Check x = -9
Log3 (-9+6) + log3 (3*-9) = 4
Each of the logs gives an i, this solution doesn’t check. -9 is not a solution
Check x = 3
Log3(3+6) + log3 (3*3) = 4
2 + 2 = 4, this solution check and x = 3 is the only solution to this problem.
Answer: x = 3
(no need to mention the x =-9 as it doesn’t check
It would be okay to write: Answer: x = 3, x = -9 is extraneous)