Exponential equation solving worksheet with logarithms for math education.

Here is the step-by-step solution to each of the 14 exponential equations using logarithms. All answers are rounded to the nearest ten-thousandth (i.e., 4 decimal places).

---

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

Add 1 to both sides:
\[
3e^{-2m+4} = 16
\]

Divide by 3:
\[
e^{-2m+4} = \frac{16}{3}
\]

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

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

Calculate:
\[
\ln(16/3) \approx \ln(5.3333) \approx 1.67398
\]
\[
m = \frac{4 - 1.67398}{2} = \frac{2.32602}{2} \approx 1.1630
\]

✔ Answer: \( \boxed{1.1630} \)

---

2) \( 12^p = 16 \)

Take log base 10 or natural log:
\[
\ln(12^p) = \ln(16)
\Rightarrow p \ln(12) = \ln(16)
\Rightarrow p = \frac{\ln(16)}{\ln(12)}
\]

Calculate:
\[
\ln(16) \approx 2.7726, \quad \ln(12) \approx 2.4849
\Rightarrow p \approx \frac{2.7726}{2.4849} \approx 1.1158
\]

✔ Answer: \( \boxed{1.1158} \)

---

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

Divide by 5:
\[
6^{3m} = \frac{17}{5} = 3.4
\]

Take log:
\[
\ln(6^{3m}) = \ln(3.4)
\Rightarrow 3m \ln(6) = \ln(3.4)
\Rightarrow m = \frac{\ln(3.4)}{3 \ln(6)}
\]

Calculate:
\[
\ln(3.4) \approx 1.2238, \quad \ln(6) \approx 1.7918
\Rightarrow m = \frac{1.2238}{3 \cdot 1.7918} = \frac{1.2238}{5.3754} \approx 0.2277
\]

✔ Answer: \( \boxed{0.2277} \)

---

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

Add 10:
\[
8^{x+10} = 28
\]

Take log:
\[
\ln(8^{x+10}) = \ln(28)
\Rightarrow (x+10)\ln(8) = \ln(28)
\Rightarrow x + 10 = \frac{\ln(28)}{\ln(8)}
\Rightarrow x = \frac{\ln(28)}{\ln(8)} - 10
\]

Calculate:
\[
\ln(28) \approx 3.3322, \quad \ln(8) \approx 2.0794
\Rightarrow \frac{3.3322}{2.0794} \approx 1.6025
\Rightarrow x \approx 1.6025 - 10 = -8.3975
\]

✔ Answer: \( \boxed{-8.3975} \)

---

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

Divide by 2:
\[
3^{3p} = 9.5
\]

Take log:
\[
\ln(3^{3p}) = \ln(9.5)
\Rightarrow 3p \ln(3) = \ln(9.5)
\Rightarrow p = \frac{\ln(9.5)}{3 \ln(3)}
\]

Calculate:
\[
\ln(9.5) \approx 2.2513, \quad \ln(3) \approx 1.0986
\Rightarrow p = \frac{2.2513}{3 \cdot 1.0986} = \frac{2.2513}{3.2958} \approx 0.6831
\]

✔ Answer: \( \boxed{0.6831} \)

---

6) \( 15^d = 20 \)

Take log:
\[
d = \frac{\ln(20)}{\ln(15)}
\]

Calculate:
\[
\ln(20) \approx 2.9957, \quad \ln(15) \approx 2.7081
\Rightarrow d \approx \frac{2.9957}{2.7081} \approx 1.1062
\]

✔ Answer: \( \boxed{1.1062} \)

---

7) \( 9^{2g+6} - 3 = 21 \)

Add 3:
\[
9^{2g+6} = 24
\]

Take log:
\[
\ln(9^{2g+6}) = \ln(24)
\Rightarrow (2g+6)\ln(9) = \ln(24)
\Rightarrow 2g + 6 = \frac{\ln(24)}{\ln(9)}
\Rightarrow g = \frac{1}{2} \left( \frac{\ln(24)}{\ln(9)} - 6 \right)
\]

Calculate:
\[
\ln(24) \approx 3.1781, \quad \ln(9) \approx 2.1972
\Rightarrow \frac{3.1781}{2.1972} \approx 1.4464
\Rightarrow g = \frac{1}{2}(1.4464 - 6) = \frac{1}{2}(-4.5536) \approx -2.2768
\]

✔ Answer: \( \boxed{-2.2768} \)

---

8) \( 15^{q+9} + 8 = 22 \)

Subtract 8:
\[
15^{q+9} = 14
\]

Take log:
\[
(q+9)\ln(15) = \ln(14)
\Rightarrow q + 9 = \frac{\ln(14)}{\ln(15)}
\Rightarrow q = \frac{\ln(14)}{\ln(15)} - 9
\]

Calculate:
\[
\ln(14) \approx 2.6391, \quad \ln(15) \approx 2.7081
\Rightarrow \frac{2.6391}{2.7081} \approx 0.9745
\Rightarrow q \approx 0.9745 - 9 = -8.0255
\]

✔ Answer: \( \boxed{-8.0255} \)

---

9) \( e^{d+4} + 8 = 23 \)

Subtract 8:
\[
e^{d+4} = 15
\]

Take natural log:
\[
d + 4 = \ln(15)
\Rightarrow d = \ln(15) - 4
\]

Calculate:
\[
\ln(15) \approx 2.7081
\Rightarrow d \approx 2.7081 - 4 = -1.2919
\]

✔ Answer: \( \boxed{-1.2919} \)

---

10) \( 4^b = 24 \)

Take log:
\[
b = \frac{\ln(24)}{\ln(4)}
\]

Calculate:
\[
\ln(24) \approx 3.1781, \quad \ln(4) \approx 1.3863
\Rightarrow b \approx \frac{3.1781}{1.3863} \approx 2.2925
\]

✔ Answer: \( \boxed{2.2925} \)

---

11) \( 4^{4y+8} - 1 = 25 \)

Add 1:
\[
4^{4y+8} = 26
\]

Take log:
\[
(4y+8)\ln(4) = \ln(26)
\Rightarrow 4y + 8 = \frac{\ln(26)}{\ln(4)}
\Rightarrow y = \frac{1}{4} \left( \frac{\ln(26)}{\ln(4)} - 8 \right)
\]

Calculate:
\[
\ln(26) \approx 3.2581, \quad \ln(4) \approx 1.3863
\Rightarrow \frac{3.2581}{1.3863} \approx 2.3499
\Rightarrow y = \frac{1}{4}(2.3499 - 8) = \frac{1}{4}(-5.6501) \approx -1.4125
\]

✔ Answer: \( \boxed{-1.4125} \)

---

12) \( 16^{a+1} + 8 = 26 \)

Subtract 8:
\[
16^{a+1} = 18
\]

Take log:
\[
(a+1)\ln(16) = \ln(18)
\Rightarrow a + 1 = \frac{\ln(18)}{\ln(16)}
\Rightarrow a = \frac{\ln(18)}{\ln(16)} - 1
\]

Calculate:
\[
\ln(18) \approx 2.8904, \quad \ln(16) \approx 2.7726
\Rightarrow \frac{2.8904}{2.7726} \approx 1.0425
\Rightarrow a \approx 1.0425 - 1 = 0.0425
\]

✔ Answer: \( \boxed{0.0425} \)

---

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

Add 4:
\[
e^{d+6} = 31
\]

Take natural log:
\[
d + 6 = \ln(31)
\Rightarrow d = \ln(31) - 6
\]

Calculate:
\[
\ln(31) \approx 3.4340
\Rightarrow d \approx 3.4340 - 6 = -2.5660
\]

✔ Answer: \( \boxed{-2.5660} \)

---

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

Subtract 6:
\[
4e^{-4d+3} = 22
\]

Divide by 4:
\[
e^{-4d+3} = 5.5
\]

Take natural log:
\[
-4d + 3 = \ln(5.5)
\Rightarrow -4d = \ln(5.5) - 3
\Rightarrow d = \frac{3 - \ln(5.5)}{4}
\]

Calculate:
\[
\ln(5.5) \approx 1.7047
\Rightarrow d = \frac{3 - 1.7047}{4} = \frac{1.2953}{4} \approx 0.3238
\]

✔ Answer: \( \boxed{0.3238} \)

---

## ✔ Final Answers Summary:

1) 1.1630
2) 1.1158
3) 0.2277
4) -8.3975
5) 0.6831
6) 1.1062
7) -2.2768
8) -8.0255
9) -1.2919
10) 2.2925
11) -1.4125
12) 0.0425
13) -2.5660
14) 0.3238

Let me know if you'd like to see any step explained further!