Algebra 2 Worksheets | Exponential and Logarithmic Functions ... - Free Printable

Let's solve each of these exponential equations using logarithms. We'll go step by step and use logarithmic properties to isolate the variable.

---

1) $ 3e^{-2m+4} - 1 = 15 $

Step 1: Add 1 to both sides
$$
3e^{-2m+4} = 16
$$

Step 2: Divide both sides by 3
$$
e^{-2m+4} = \frac{16}{3}
$$

Step 3: Take natural log (ln) of both sides
$$
\ln(e^{-2m+4}) = \ln\left(\frac{16}{3}\right)
$$

$$
-2m + 4 = \ln\left(\frac{16}{3}\right)
$$

Step 4: Solve for $ m $
$$
-2m = \ln\left(\frac{16}{3}\right) - 4
$$

$$
m = \frac{4 - \ln\left(\frac{16}{3}\right)}{2}
$$

Calculate:
$$
\ln\left(\frac{16}{3}\right) \approx \ln(5.333) \approx 1.674
$$

$$
m = \frac{4 - 1.674}{2} = \frac{2.326}{2} = 1.1630
$$

✔ Answer: $ m \approx 1.1630 $

---

2) $ 12^p = 16 $

Take log of both sides (can use common or natural log):

$$
\log(12^p) = \log(16)
\Rightarrow p \log(12) = \log(16)
$$

$$
p = \frac{\log(16)}{\log(12)}
$$

$$
\log(16) = \log(2^4) = 4\log(2) \approx 4(0.3010) = 1.204
$$

$$
\log(12) = \log(3 \cdot 4) = \log(3) + \log(4) \approx 0.4771 + 0.6021 = 1.0792
$$

$$
p \approx \frac{1.204}{1.0792} \approx 1.1163
$$

✔ Answer: $ p \approx 1.1163 $

---

3) $ 5 \cdot 6^{-3m} = 17 $

Divide both sides by 5:

$$
6^{-3m} = \frac{17}{5} = 3.4
$$

Take log:

$$
\log(6^{-3m}) = \log(3.4)
\Rightarrow -3m \log(6) = \log(3.4)
$$

$$
m = -\frac{\log(3.4)}{3 \log(6)}
$$

$$
\log(3.4) \approx 0.5315, \quad \log(6) \approx 0.7782
$$

$$
m \approx -\frac{0.5315}{3 \cdot 0.7782} = -\frac{0.5315}{2.3346} \approx -0.2276
$$

✔ Answer: $ m \approx -0.2276 $

---

4) $ 8^{x+10} - 10 = 18 $

Add 10:

$$
8^{x+10} = 28
$$

Take log:

$$
\log(8^{x+10}) = \log(28)
\Rightarrow (x+10)\log(8) = \log(28)
$$

$$
x + 10 = \frac{\log(28)}{\log(8)}
$$

$$
\log(28) \approx 1.4472, \quad \log(8) = \log(2^3) = 3\log(2) \approx 3(0.3010) = 0.9030
$$

$$
x + 10 = \frac{1.4472}{0.9030} \approx 1.5999
\Rightarrow x \approx 1.5999 - 10 = -8.4001
$$

✔ Answer: $ x \approx -8.4001 $

---

5) $ 2 \cdot 3^{3p} = 19 $

Divide by 2:

$$
3^{3p} = \frac{19}{2} = 9.5
$$

Take log:

$$
\log(3^{3p}) = \log(9.5)
\Rightarrow 3p \log(3) = \log(9.5)
$$

$$
p = \frac{\log(9.5)}{3 \log(3)}
$$

$$
\log(9.5) \approx 0.9777, \quad \log(3) \approx 0.4771
$$

$$
p \approx \frac{0.9777}{3 \cdot 0.4771} = \frac{0.9777}{1.4313} \approx 0.6830
$$

✔ Answer: $ p \approx 0.6830 $

---

6) $ 15^d = 20 $

Take log:

$$
\log(15^d) = \log(20)
\Rightarrow d \log(15) = \log(20)
$$

$$
d = \frac{\log(20)}{\log(15)}
$$

$$
\log(20) = \log(2 \cdot 10) = \log(2) + 1 \approx 0.3010 + 1 = 1.3010
$$

$$
\log(15) = \log(3 \cdot 5) = \log(3) + \log(5) \approx 0.4771 + 0.6990 = 1.1761
$$

$$
d \approx \frac{1.3010}{1.1761} \approx 1.1066
$$

✔ Answer: $ d \approx 1.1066 $

---

7) $ 9^{2q+6} - 3 = 21 $

Add 3:

$$
9^{2q+6} = 24
$$

Note: $ 9 = 3^2 $, so $ 9^{2q+6} = (3^2)^{2q+6} = 3^{2(2q+6)} = 3^{4q+12} $

But we can just take log directly:

$$
\log(9^{2q+6}) = \log(24)
\Rightarrow (2q+6)\log(9) = \log(24)
$$

$$
2q + 6 = \frac{\log(24)}{\log(9)}
$$

$$
\log(24) \approx 1.3802, \quad \log(9) = \log(3^2) = 2\log(3) \approx 2(0.4771) = 0.9542
$$

$$
2q + 6 = \frac{1.3802}{0.9542} \approx 1.4457
\Rightarrow 2q = 1.4457 - 6 = -4.5543
\Rightarrow q = -2.2772
$$

✔ Answer: $ q \approx -2.2772 $

---

8) $ 15^{n+9} + 8 = 22 $

Subtract 8:

$$
15^{n+9} = 14
$$

Take log:

$$
(n+9)\log(15) = \log(14)
\Rightarrow n + 9 = \frac{\log(14)}{\log(15)}
$$

$$
\log(14) \approx 1.1461, \quad \log(15) \approx 1.1761
$$

$$
n + 9 = \frac{1.1461}{1.1761} \approx 0.9745
\Rightarrow n \approx 0.9745 - 9 = -8.0255
$$

✔ Answer: $ n \approx -8.0255 $

---

9) $ e^{d-4} + 8 = 23 $

Subtract 8:

$$
e^{d-4} = 15
$$

Take natural log:

$$
\ln(e^{d-4}) = \ln(15)
\Rightarrow d - 4 = \ln(15)
\Rightarrow d = \ln(15) + 4
$$

$$
\ln(15) \approx 2.7081
\Rightarrow d \approx 2.7081 + 4 = 6.7081
$$

✔ Answer: $ d \approx 6.7081 $

---

10) $ 4^b = 24 $

Take log:

$$
\log(4^b) = \log(24)
\Rightarrow b \log(4) = \log(24)
\Rightarrow b = \frac{\log(24)}{\log(4)}
$$

$$
\log(24) \approx 1.3802, \quad \log(4) = \log(2^2) = 2\log(2) \approx 0.6021
$$

$$
b \approx \frac{1.3802}{0.6021} \approx 2.2914
$$

✔ Answer: $ b \approx 2.2914 $

---

11) $ 4^{4y+9} - 1 = 25 $

Add 1:

$$
4^{4y+9} = 26
$$

Take log:

$$
(4y+9)\log(4) = \log(26)
\Rightarrow 4y + 9 = \frac{\log(26)}{\log(4)}
$$

$$
\log(26) \approx 1.4149, \quad \log(4) \approx 0.6021
$$

$$
4y + 9 = \frac{1.4149}{0.6021} \approx 2.3493
\Rightarrow 4y = 2.3493 - 9 = -6.6507
\Rightarrow y = -1.6627
$$

✔ Answer: $ y \approx -1.6627 $

---

12) $ 16^{p+1} + 8 = 26 $

Subtract 8:

$$
16^{p+1} = 18
$$

Take log:

$$
(p+1)\log(16) = \log(18)
\Rightarrow p + 1 = \frac{\log(18)}{\log(16)}
$$

$$
\log(18) \approx 1.2553, \quad \log(16) = \log(2^4) = 4\log(2) \approx 1.2041
$$

$$
p + 1 = \frac{1.2553}{1.2041} \approx 1.0425
\Rightarrow p \approx 0.0425
$$

✔ Answer: $ p \approx 0.0425 $

---

13) $ e^{d+6} - 4 = 27 $

Add 4:

$$
e^{d+6} = 31
$$

Take ln:

$$
\ln(e^{d+6}) = \ln(31)
\Rightarrow d + 6 = \ln(31)
\Rightarrow d = \ln(31) - 6
$$

$$
\ln(31) \approx 3.4339
\Rightarrow d \approx 3.4339 - 6 = -2.5661
$$

✔ Answer: $ d \approx -2.5661 $

---

14) $ 4e^{4d-3} + 6 = 28 $

Subtract 6:

$$
4e^{4d-3} = 22
\Rightarrow e^{4d-3} = \frac{22}{4} = 5.5
$$

Take ln:

$$
\ln(e^{4d-3}) = \ln(5.5)
\Rightarrow 4d - 3 = \ln(5.5)
\Rightarrow 4d = \ln(5.5) + 3
$$

$$
\ln(5.5) \approx 1.7047
\Rightarrow 4d = 1.7047 + 3 = 4.7047
\Rightarrow d = \frac{4.7047}{4} = 1.1762
$$

✔ Answer: $ d \approx 1.1762 $

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## ✔ Final Answers (Rounded to Nearest Ten-Thousandth):

| Problem | Answer |
|--------|--------|
| 1 | $ m \approx 1.1630 $ |
| 2 | $ p \approx 1.1163 $ |
| 3 | $ m \approx -0.2276 $ |
| 4 | $ x \approx -8.4001 $ |
| 5 | $ p \approx 0.6830 $ |
| 6 | $ d \approx 1.1066 $ |
| 7 | $ q \approx -2.2772 $ |
| 8 | $ n \approx -8.0255 $ |
| 9 | $ d \approx 6.7081 $ |
| 10 | $ b \approx 2.2914 $ |
| 11 | $ y \approx -1.6627 $ |
| 12 | $ p \approx 0.0425 $ |
| 13 | $ d \approx -2.5661 $ |
| 14 | $ d \approx 1.1762 $ |

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