Algebraic exponent equations worksheet with 30 problems.

To solve the problems in the image, we need to use the properties of exponents and logarithms. Let's go through each problem step by step.

Problem 7: \( 15^{2x} = 125 \)

1. Express 125 as a power of 5:
\[
125 = 5^3
\]
2. Rewrite the equation:
\[
15^{2x} = 5^3
\]
3. Since the bases are not the same, we cannot directly equate the exponents. However, we can take the logarithm of both sides:
\[
\log(15^{2x}) = \log(5^3)
\]
4. Use the logarithmic property \(\log(a^b) = b \log(a)\):
\[
2x \log(15) = 3 \log(5)
\]
5. Solve for \(x\):
\[
x = \frac{3 \log(5)}{2 \log(15)}
\]

Problem 8: \( 16^{x-2} = 32^{x+1} \)

1. Express 16 and 32 as powers of 2:
\[
16 = 2^4 \quad \text{and} \quad 32 = 2^5
\]
2. Rewrite the equation using these powers:
\[
(2^4)^{x-2} = (2^5)^{x+1}
\]
3. Simplify using the property \((a^m)^n = a^{mn}\):
\[
2^{4(x-2)} = 2^{5(x+1)}
\]
4. Since the bases are the same, equate the exponents:
\[
4(x-2) = 5(x+1)
\]
5. Solve for \(x\):
\[
4x - 8 = 5x + 5
\]
\[
4x - 5x = 5 + 8
\]
\[
-x = 13
\]
\[
x = -13
\]

Problem 9: \( 9^{x+2} = 27^{x+1} \)

1. Express 9 and 27 as powers of 3:
\[
9 = 3^2 \quad \text{and} \quad 27 = 3^3
\]
2. Rewrite the equation using these powers:
\[
(3^2)^{x+2} = (3^3)^{x+1}
\]
3. Simplify using the property \((a^m)^n = a^{mn}\):
\[
3^{2(x+2)} = 3^{3(x+1)}
\]
4. Since the bases are the same, equate the exponents:
\[
2(x+2) = 3(x+1)
\]
5. Solve for \(x\):
\[
2x + 4 = 3x + 3
\]
\[
2x - 3x = 3 - 4
\]
\[
-x = -1
\]
\[
x = 1
\]

Problem 10: \( 25^{x+1} = 125^{x+1} \)

1. Express 25 and 125 as powers of 5:
\[
25 = 5^2 \quad \text{and} \quad 125 = 5^3
\]
2. Rewrite the equation using these powers:
\[
(5^2)^{x+1} = (5^3)^{x+1}
\]
3. Simplify using the property \((a^m)^n = a^{mn}\):
\[
5^{2(x+1)} = 5^{3(x+1)}
\]
4. Since the bases are the same, equate the exponents:
\[
2(x+1) = 3(x+1)
\]
5. Solve for \(x\):
\[
2x + 2 = 3x + 3
\]
\[
2x - 3x = 3 - 2
\]
\[
-x = 1
\]
\[
x = -1
\]

Problem 11: \( \left(\frac{1}{4}\right)^x = 16 \)

1. Express 16 as a power of 4:
\[
16 = 4^2
\]
2. Rewrite the equation:
\[
\left(\frac{1}{4}\right)^x = 4^2
\]
3. Express \(\frac{1}{4}\) as \(4^{-1}\):
\[
(4^{-1})^x = 4^2
\]
4. Simplify using the property \((a^m)^n = a^{mn}\):
\[
4^{-x} = 4^2
\]
5. Since the bases are the same, equate the exponents:
\[
-x = 2
\]
6. Solve for \(x\):
\[
x = -2
\]

Problem 12: \( 6^{-x} = \frac{1}{36} \)

1. Express \(\frac{1}{36}\) as a power of 6:
\[
\frac{1}{36} = 6^{-2}
\]
2. Rewrite the equation:
\[
6^{-x} = 6^{-2}
\]
3. Since the bases are the same, equate the exponents:
\[
-x = -2
\]
4. Solve for \(x\):
\[
x = 2
\]

Problem 13: \( \left(\frac{1}{2}\right)^x = 2^{-x} \)

1. The equation is already in a simplified form:
\[
\left(\frac{1}{2}\right)^x = 2^{-x}
\]
2. Express \(\frac{1}{2}\) as \(2^{-1}\):
\[
(2^{-1})^x = 2^{-x}
\]
3. Simplify using the property \((a^m)^n = a^{mn}\):
\[
2^{-x} = 2^{-x}
\]
4. This equation is true for all \(x\).

Problem 14: \( \left(\frac{1}{3}\right)^x = 3^{-x} \)

1. The equation is already in a simplified form:
\[
\left(\frac{1}{3}\right)^x = 3^{-x}
\]
2. Express \(\frac{1}{3}\) as \(3^{-1}\):
\[
(3^{-1})^x = 3^{-x}
\]
3. Simplify using the property \((a^m)^n = a^{mn}\):
\[
3^{-x} = 3^{-x}
\]
4. This equation is true for all \(x\).

Problem 15: \( 3^{x-y} = \frac{1}{243} \)

1. Express \(\frac{1}{243}\) as a power of 3:
\[
\frac{1}{243} = 3^{-5}
\]
2. Rewrite the equation:
\[
3^{x-y} = 3^{-5}
\]
3. Since the bases are the same, equate the exponents:
\[
x - y = -5
\]

Problem 16: \( 5^{x-y} = \frac{1}{625} \)

1. Express \(\frac{1}{625}\) as a power of 5:
\[
\frac{1}{625} = 5^{-4}
\]
2. Rewrite the equation:
\[
5^{x-y} = 5^{-4}
\]
3. Since the bases are the same, equate the exponents:
\[
x - y = -4
\]

Problem 17: \( 2^{x-y} = \frac{1}{128} \)

1. Express \(\frac{1}{128}\) as a power of 2:
\[
\frac{1}{128} = 2^{-7}
\]
2. Rewrite the equation:
\[
2^{x-y} = 2^{-7}
\]
3. Since the bases are the same, equate the exponents:
\[
x - y = -7
\]

Problem 18: \( \left(\frac{1}{9}\right)^{\frac{x}{y}} = 3^{y-x} \)

1. Express \(\frac{1}{9}\) as \(3^{-2}\):
\[
\left(3^{-2}\right)^{\frac{x}{y}} = 3^{y-x}
\]
2. Simplify using the property \((a^m)^n = a^{mn}\):
\[
3^{-2 \cdot \frac{x}{y}} = 3^{y-x}
\]
3. Since the bases are the same, equate the exponents:
\[
-2 \cdot \frac{x}{y} = y - x
\]
4. Solve for \(x\) and \(y\):
\[
-2x = y(y - x)
\]
\[
-2x = y^2 - xy
\]
\[
xy - 2x = y^2
\]
\[
x(y - 2) = y^2
\]
\[
x = \frac{y^2}{y - 2}
\]

Final Answers

\[
\boxed{
\begin{aligned}
&7. \ x = \frac{3 \log(5)}{2 \log(15)} \\
&8. \ x = -13 \\
&9. \ x = 1 \\
&10. \ x = -1 \\
&11. \ x = -2 \\
&12. \ x = 2 \\
&13. \ x \in \mathbb{R} \\
&14. \ x \in \mathbb{R} \\
&15. \ x - y = -5 \\
&16. \ x - y = -4 \\
&17. \ x - y = -7 \\
&18. \ x = \frac{y^2}{y - 2}
\end{aligned}
}
\]