Solving Exponential and Logarithmic Equations worksheet with problems and color-by-number instructions.

Here are the step-by-step solutions for each problem on the worksheet.

1. $16^x = 64^{x-3}$
* Rewrite both bases as powers of 4: $16 = 4^2$ and $64 = 4^3$.
* The equation becomes $(4^2)^x = (4^3)^{x-3}$.
* Multiply the exponents: $4^{2x} = 4^{3(x-3)}$.
* Since the bases are the same, set the exponents equal: $2x = 3(x - 3)$.
* Distribute the 3: $2x = 3x - 9$.
* Subtract $3x$ from both sides: $-x = -9$.
* Divide by -1: $x = 9$.

2. $32^{2x+4} = 8^{-7x}$
* Rewrite both bases as powers of 2: $32 = 2^5$ and $8 = 2^3$.
* The equation becomes $(2^5)^{2x+4} = (2^3)^{-7x}$.
* Multiply the exponents: $2^{5(2x+4)} = 2^{3(-7x)}$.
* Set the exponents equal: $5(2x + 4) = -21x$.
* Distribute the 5: $10x + 20 = -21x$.
* Add $21x$ to both sides: $31x + 20 = 0$.
* Subtract 20: $31x = -20$.
* Divide by 31: $x = -\frac{20}{31}$.

3. $-4 \cdot 10^{x+12} = -48$
* Divide both sides by -4: $10^{x+12} = 12$.
* Convert to logarithmic form: $\log_{10}(12) = x + 12$.
* Subtract 12 from both sides: $x = \log(12) - 12$.
*(Note: $\log(12)$ is approx 1.079, so $x \approx -10.92$)*

4. $10 \cdot 10^{x-11} = 11$
* Divide both sides by 10: $10^{x-11} = 1.1$.
* Convert to logarithmic form: $\log_{10}(1.1) = x - 11$.
* Add 11 to both sides: $x = \log(1.1) + 11$.
*(Note: $\log(1.1)$ is approx 0.041, so $x \approx 11.04$)*

5. $20^{-4x} + 30 = 99$
* Subtract 30 from both sides: $20^{-4x} = 69$.
* Take the log of both sides: $\log(20^{-4x}) = \log(69)$.
* Bring the exponent down: $-4x \cdot \log(20) = \log(69)$.
* Divide by $-4\log(20)$: $x = \frac{\log(69)}{-4\log(20)}$.
*(Note: $x \approx -0.35$)*

6. $15^{x-9} - 2 = 73$
* Add 2 to both sides: $15^{x-9} = 75$.
* Take the log of both sides: $\log(15^{x-9}) = \log(75)$.
* Bring the exponent down: $(x - 9)\log(15) = \log(75)$.
* Divide by $\log(15)$: $x - 9 = \frac{\log(75)}{\log(15)}$.
* Add 9 to both sides: $x = \frac{\log(75)}{\log(15)} + 9$.
*(Note: $\frac{\log(75)}{\log(15)} \approx 1.68$, so $x \approx 10.68$)*

7. $\ln(-2x + 6) = \ln(12)$
* Since both sides are natural logs ($\ln$), you can drop the logs: $-2x + 6 = 12$.
* Subtract 6 from both sides: $-2x = 6$.
* Divide by -2: $x = -3$.
*(Check: $-2(-3)+6 = 12$, which is positive, so the solution is valid.)*

8. $(\log_9(x+3)) - (\log_9(12)) = \log_9(10)$
* Use the quotient rule for logs on the left side: $\log_9(\frac{x+3}{12}) = \log_9(10)$.
* Drop the logs: $\frac{x+3}{12} = 10$.
* Multiply by 12: $x + 3 = 120$.
* Subtract 3: $x = 117$.

9. $(\log_8(9x)) - (\log_8(3)) = 4$
* Use the quotient rule for logs on the left side: $\log_8(\frac{9x}{3}) = 4$.
* Simplify the fraction: $\log_8(3x) = 4$.
* Convert to exponential form: $8^4 = 3x$.
* Calculate $8^4$: $4096 = 3x$.
* Divide by 3: $x = \frac{4096}{3}$.

10. $(\log_7(x+3)) - (\log_7(6)) = 1$
* Use the quotient rule for logs on the left side: $\log_7(\frac{x+3}{6}) = 1$.
* Convert to exponential form: $7^1 = \frac{x+3}{6}$.
* So, $7 = \frac{x+3}{6}$.
* Multiply by 6: $42 = x + 3$.
* Subtract 3: $x = 39$.

11. $\log_3(x) + \log_3(x-13) = \log_3(30)$
* Use the product rule for logs on the left side: $\log_3(x(x-13)) = \log_3(30)$.
* Drop the logs: $x(x - 13) = 30$.
* Distribute $x$: $x^2 - 13x = 30$.
* Set to zero: $x^2 - 13x - 30 = 0$.
* Factor the quadratic: Find two numbers that multiply to -30 and add to -13. These are -15 and +2.
* $(x - 15)(x + 2) = 0$.
* Possible solutions: $x = 15$ or $x = -2$.
* Check validity: You cannot take the log of a negative number.
* If $x = -2$, $\log_3(-2)$ is undefined. So, reject -2.
* If $x = 15$, $\log_3(15)$ and $\log_3(2)$ are both defined.
* Final Answer: $x = 15$.

Final Answer:
1. $x = 9$
2. $x = -\frac{20}{31}$
3. $x = \log(12) - 12$
4. $x = \log(1.1) + 11$
5. $x = \frac{\log(69)}{-4\log(20)}$
6. $x = \frac{\log(75)}{\log(15)} + 9$
7. $x = -3$
8. $x = 117$
9. $x = \frac{4096}{3}$
10. $x = 39$
11. $x = 15$