Exponential equations worksheet with problems labeled a to z.

To solve the given problems, we will use properties of exponents and logarithms. Let's go through each problem step by step.

Part a) \( 3^{2\lambda} = 9 \)

1. Rewrite 9 as a power of 3:
\[
9 = 3^2
\]
2. Equate the exponents:
\[
3^{2\lambda} = 3^2 \implies 2\lambda = 2
\]
3. Solve for \(\lambda\):
\[
\lambda = 1
\]

Part b) \( 5^{3\lambda+2} = 25^{\lambda-1} \)

1. Rewrite 25 as a power of 5:
\[
25 = 5^2
\]
2. Substitute and equate the exponents:
\[
5^{3\lambda+2} = (5^2)^{\lambda-1} = 5^{2(\lambda-1)}
\]
\[
3\lambda + 2 = 2(\lambda - 1)
\]
3. Simplify and solve for \(\lambda\):
\[
3\lambda + 2 = 2\lambda - 2
\]
\[
3\lambda - 2\lambda = -2 - 2
\]
\[
\lambda = -4
\]

Part c) \( 2^{3-\lambda} = 4^{2-\lambda} \)

1. Rewrite 4 as a power of 2:
\[
4 = 2^2
\]
2. Substitute and equate the exponents:
\[
2^{3-\lambda} = (2^2)^{2-\lambda} = 2^{2(2-\lambda)}
\]
\[
3 - \lambda = 2(2 - \lambda)
\]
3. Simplify and solve for \(\lambda\):
\[
3 - \lambda = 4 - 2\lambda
\]
\[
3 - 4 = -2\lambda + \lambda
\]
\[
-1 = -\lambda
\]
\[
\lambda = 1
\]

Part d) \( 8^{-\lambda} = 2^{\lambda-8} \)

1. Rewrite 8 as a power of 2:
\[
8 = 2^3
\]
2. Substitute and equate the exponents:
\[
(2^3)^{-\lambda} = 2^{\lambda-8}
\]
\[
2^{-3\lambda} = 2^{\lambda-8}
\]
3. Equate the exponents:
\[
-3\lambda = \lambda - 8
\]
4. Solve for \(\lambda\):
\[
-3\lambda - \lambda = -8
\]
\[
-4\lambda = -8
\]
\[
\lambda = 2
\]

Part e) \( \left( \frac{1}{4} \right)^{x-1} = 4^{2-3x} \)

1. Rewrite \(\frac{1}{4}\) as a power of 4:
\[
\frac{1}{4} = 4^{-1}
\]
2. Substitute and equate the exponents:
\[
(4^{-1})^{x-1} = 4^{2-3x}
\]
\[
4^{-(x-1)} = 4^{2-3x}
\]
3. Equate the exponents:
\[
-(x-1) = 2 - 3x
\]
4. Simplify and solve for \(x\):
\[
-x + 1 = 2 - 3x
\]
\[
-x + 3x = 2 - 1
\]
\[
2x = 1
\]
\[
x = \frac{1}{2}
\]

Part f) \( 5^{3x+2} = 25^{\lambda-1} \)

This is the same as part b). The solution is:
\[
\lambda = -4
\]

Part g) \( 4^x = 2^{x+1} \)

1. Rewrite 4 as a power of 2:
\[
4 = 2^2
\]
2. Substitute and equate the exponents:
\[
(2^2)^x = 2^{x+1}
\]
\[
2^{2x} = 2^{x+1}
\]
3. Equate the exponents:
\[
2x = x + 1
\]
4. Solve for \(x\):
\[
2x - x = 1
\]
\[
x = 1
\]

Part h) \( 8^{2x+1} = \left( \frac{1}{16} \right)^{3-2x} \)

1. Rewrite 8 and \(\frac{1}{16}\) as powers of 2:
\[
8 = 2^3, \quad \frac{1}{16} = 2^{-4}
\]
2. Substitute and equate the exponents:
\[
(2^3)^{2x+1} = (2^{-4})^{3-2x}
\]
\[
2^{3(2x+1)} = 2^{-4(3-2x)}
\]
3. Equate the exponents:
\[
3(2x+1) = -4(3-2x)
\]
4. Simplify and solve for \(x\):
\[
6x + 3 = -12 + 8x
\]
\[
6x - 8x = -12 - 3
\]
\[
-2x = -15
\]
\[
x = \frac{15}{2}
\]

Part i) \( \left( \frac{1}{2} \right)^{3x} = 2^{-x-1} \)

1. Rewrite \(\frac{1}{2}\) as a power of 2:
\[
\frac{1}{2} = 2^{-1}
\]
2. Substitute and equate the exponents:
\[
(2^{-1})^{3x} = 2^{-x-1}
\]
\[
2^{-3x} = 2^{-x-1}
\]
3. Equate the exponents:
\[
-3x = -x - 1
\]
4. Solve for \(x\):
\[
-3x + x = -1
\]
\[
-2x = -1
\]
\[
x = \frac{1}{2}
\]

Part j) \( 2^{-3x+1} = \left( \frac{1}{2} \right)^{-x-1} \)

1. Rewrite \(\frac{1}{2}\) as a power of 2:
\[
\frac{1}{2} = 2^{-1}
\]
2. Substitute and equate the exponents:
\[
2^{-3x+1} = (2^{-1})^{-x-1}
\]
\[
2^{-3x+1} = 2^{x+1}
\]
3. Equate the exponents:
\[
-3x + 1 = x + 1
\]
4. Solve for \(x\):
\[
-3x - x = 1 - 1
\]
\[
-4x = 0
\]
\[
x = 0
\]

Part k) \( 2^{x-1} = \left( \frac{1}{4} \right)^{2-4x} \)

1. Rewrite \(\frac{1}{4}\) as a power of 2:
\[
\frac{1}{4} = 2^{-2}
\]
2. Substitute and equate the exponents:
\[
2^{x-1} = (2^{-2})^{2-4x}
\]
\[
2^{x-1} = 2^{-2(2-4x)}
\]
3. Equate the exponents:
\[
x - 1 = -2(2 - 4x)
\]
4. Simplify and solve for \(x\):
\[
x - 1 = -4 + 8x
\]
\[
x - 8x = -4 + 1
\]
\[
-7x = -3
\]
\[
x = \frac{3}{7}
\]

Part l) \( \left( \frac{1}{27} \right)^{4-x} = 9^{2x} \)

1. Rewrite \(\frac{1}{27}\) and 9 as powers of 3:
\[
\frac{1}{27} = 3^{-3}, \quad 9 = 3^2
\]
2. Substitute and equate the exponents:
\[
(3^{-3})^{4-x} = (3^2)^{2x}
\]
\[
3^{-3(4-x)} = 3^{4x}
\]
3. Equate the exponents:
\[
-3(4-x) = 4x
\]
4. Simplify and solve for \(x\):
\[
-12 + 3x = 4x
\]
\[
-12 = 4x - 3x
\]
\[
-12 = x
\]
\[
x = -12
\]

Part m) \( \left( \frac{3}{5} \right)^x = \left( \frac{5}{3} \right)^3 \)

1. Rewrite \(\frac{5}{3}\) as a reciprocal of \(\frac{3}{5}\):
\[
\frac{5}{3} = \left( \frac{3}{5} \right)^{-1}
\]
2. Substitute and equate the exponents:
\[
\left( \frac{3}{5} \right)^x = \left( \frac{3}{5} \right)^{-3}
\]
3. Equate the exponents:
\[
x = -3
\]

Part n) \( 2^{3\lambda-4} = \left( \frac{1}{8} \right)^{x+1} \)

1. Rewrite \(\frac{1}{8}\) as a power of 2:
\[
\frac{1}{8} = 2^{-3}
\]
2. Substitute and equate the exponents:
\[
2^{3\lambda-4} = (2^{-3})^{x+1}
\]
\[
2^{3\lambda-4} = 2^{-3(x+1)}
\]
3. Equate the exponents:
\[
3\lambda - 4 = -3(x + 1)
\]
4. Simplify and solve for \(\lambda\):
\[
3\lambda - 4 = -3x - 3
\]
\[
3\lambda = -3x - 3 + 4
\]
\[
3\lambda = -3x + 1
\]
\[
\lambda = \frac{-3x + 1}{3}
\]

Part o) \( 4^{1-x} = \left( \frac{1}{4} \right)^{2x-3} \)

1. Rewrite 4 and \(\frac{1}{4}\) as powers of 2:
\[
4 = 2^2, \quad \frac{1}{4} = 2^{-2}
\]
2. Substitute and equate the exponents:
\[
(2^2)^{1-x} = (2^{-2})^{2x-3}
\]
\[
2^{2(1-x)} = 2^{-2(2x-3)}
\]
3. Equate the exponents:
\[
2(1-x) = -2(2x-3)
\]
4. Simplify and solve for \(x\):
\[
2 - 2x = -4x + 6
\]
\[
2x - 2x = 6 - 2
\]
\[
2x = 4
\]
\[
x = 2
\]

Part p) \( 10^x = 0.1 \times 1,000^{x-1} \)

1. Rewrite 0.1 and 1,000 as powers of 10:
\[
0.1 = 10^{-1}, \quad 1,000 = 10^3
\]
2. Substitute and simplify:
\[
10^x = 10^{-1} \times (10^3)^{x-1}
\]
\[
10^x = 10^{-1} \times 10^{3(x-1)}
\]
\[
10^x = 10^{-1 + 3(x-1)}
\]
\[
10^x = 10^{-1 + 3x - 3}
\]
\[
10^x = 10^{3x - 4}
\]
3. Equate the exponents:
\[
x = 3x - 4
\]
4. Solve for \(x\):
\[
x - 3x = -4
\]
\[
-2x = -4
\]
\[
x = 2
\]

Part q) \( 27 \times 27^{2x-3} = 81^{3x-5} \)

1. Rewrite 27 and 81 as powers of 3:
\[
27 = 3^3, \quad 81 = 3^4
\]
2. Substitute and simplify:
\[
(3^3) \times (3^3)^{2x-3} = (3^4)^{3x-5}
\]
\[
3^3 \times 3^{3(2x-3)} = 3^{4(3x-5)}
\]
\[
3^{3 + 3(2x-3)} = 3^{4(3x-5)}
\]
3. Equate the exponents:
\[
3 + 3(2x-3) = 4(3x-5)
\]
4. Simplify and solve for \(x\):
\[
3 + 6x - 9 = 12x - 20
\]
\[
6x - 6 = 12x - 20
\]
\[
6x - 12x = -20 + 6
\]
\[
-6x = -14
\]
\[
x = \frac{14}{6} = \frac{7}{3}
\]

Part r) \( 4 \times 2^{x+1} = \left( \frac{1}{8} \right)^{2x-3} \)

1. Rewrite 4 and \(\frac{1}{8}\) as powers of 2:
\[
4 = 2^2, \quad \frac{1}{8} = 2^{-3}
\]
2. Substitute and simplify:
\[
2^2 \times 2^{x+1} = (2^{-3})^{2x-3}
\]
\[
2^{2 + (x+1)} = 2^{-3(2x-3)}
\]
\[
2^{x+3} = 2^{-6x + 9}
\]
3. Equate the exponents:
\[
x + 3 = -6x + 9
\]
4. Solve for \(x\):
\[
x + 6x = 9 - 3
\]
\[
7x = 6
\]
\[
x = \frac{6}{7}
\]

Part s) \( 4^x = 16^{2-x} \)

1. Rewrite 16 as a power of 4:
\[
16 = 4^2
\]
2. Substitute and equate the exponents:
\[
4^x = (4^2)^{2-x}
\]
\[
4^x = 4^{2(2-x)}
\]
3. Equate the exponents:
\[
x = 2(2-x)
\]
4. Solve for \(x\):
\[
x = 4 - 2x
\]
\[
x + 2x = 4
\]
\[
3x = 4
\]
\[
x = \frac{4}{3}
\]

Part t) \( 4^{x-2} = 0.125 \)

1. Rewrite 0.125 as a power of 4:
\[
0.125 = \frac{1}{8} = 2^{-3} = (2^2)^{-\frac{3}{2}} = 4^{-\frac{3}{2}}
\]
2. Equate the exponents:
\[
4^{x-2} = 4^{-\frac{3}{2}}
\]
3. Solve for \(x\):
\[
x - 2 = -\frac{3}{2}
\]
\[
x = 2 - \frac{3}{2}
\]
\[
x = \frac{4}{2} - \frac{3}{2}
\]
\[
x = \frac{1}{2}
\]

Part u) \( \left( \frac{1}{8} \right)^{-x} = \left( \frac{1}{32} \right)^{1-x} \)

1. Rewrite \(\frac{1}{8}\) and \(\frac{1}{32}\) as powers of 2:
\[
\frac{1}{8} = 2^{-3}, \quad \frac{1}{32} = 2^{-5}
\]
2. Substitute and equate the exponents:
\[
(2^{-3})^{-x} = (2^{-5})^{1-x}
\]
\[
2^{3x} = 2^{-5(1-x)}
\]
3. Equate the exponents:
\[
3x = -5(1-x)
\]
4. Simplify and solve for \(x\):
\[
3x = -5 + 5x
\]
\[
3x - 5x = -5
\]
\[
-2x = -5
\]
\[
x = \frac{5}{2}
\]

Part v) \( \left( \frac{1}{243} \right)^{2x} = 81^{1-x} \)

1. Rewrite \(\frac{1}{243}\) and 81 as powers of 3:
\[
\frac{1}{243} = 3^{-5}, \quad 81 = 3^4
\]
2. Substitute and equate the exponents:
\[
(3^{-5})^{2x} = (3^4)^{1-x}
\]
\[
3^{-10x} = 3^{4(1-x)}
\]
3. Equate the exponents:
\[
-10x = 4(1-x)
\]
4. Simplify and solve for \(x\):
\[
-10x = 4 - 4x
\]
\[
-10x + 4x = 4
\]
\[
-6x = 4
\]
\[
x = -\frac{2}{3}
\]

Part w) \( \left( \frac{1}{125} \right)^{-3x-1} = 25^{-x-1} \)

1. Rewrite \(\frac{1}{125}\) and 25 as powers of 5:
\[
\frac{1}{125} = 5^{-3}, \quad 25 = 5^2
\]
2. Substitute and equate the exponents:
\[
(5^{-3})^{-3x-1} = (5^2)^{-x-1}
\]
\[
5^{3(3x+1)} = 5^{-2(x+1)}
\]
3. Equate the exponents:
\[
3(3x+1) = -2(x+1)
\]
4. Simplify and solve for \(x\):
\[
9x + 3 = -2x - 2
\]
\[
9x + 2x = -2 - 3
\]
\[
11x = -5
\]
\[
x = -\frac{5}{11}
\]

Part x) \( 3^{x-2} = \left( \frac{1}{3} \right)^{-2x} \)

1. Rewrite \(\frac{1}{3}\) as a power of 3:
\[
\frac{1}{3} = 3^{-1}
\]
2. Substitute and equate the exponents:
\[
3^{x-2} = (3^{-1})^{-2x}
\]
\[
3^{x-2} = 3^{2x}
\]
3. Equate the exponents:
\[
x - 2 = 2x
\]
4. Solve for \(x\):
\[
x - 2x = 2
\]
\[
-x = 2
\]
\[
x = -2
\]

Part y) \( \left( \frac{1}{3} \right)^{1-x} = 243^{-x-1} \)

1. Rewrite \(\frac{1}{3}\) and 243 as powers of 3:
\[
\frac{1}{3} = 3^{-1}, \quad 243 = 3^5
\]
2. Substitute and equate the exponents:
\[
(3^{-1})^{1-x} = (3^5)^{-x-1}
\]
\[
3^{-(1-x)} = 3^{-5(x+1)}
\]
3. Equate the exponents:
\[
-(1-x) = -5(x+1)
\]
4. Simplify and solve for \(x\):
\[
-1 + x = -5x - 5
\]
\[
x + 5x = -5 + 1
\]
\[
6x = -4
\]
\[
x = -\frac{2}{3}
\]

Part z) \( 5^{3x+2-(1-x)} = 125^{2x-1-x} \)

1. Simplify the exponents:
\[
5^{3x+2-(1-x)} = 5^{3x+2-1+x} = 5^{4x+1}
\]
\[
125^{2x-1-x} = (5^3)^{x-1} = 5^{3(x-1)} = 5^{3x-3}
\]
2. Equate the exponents:
\[
4x + 1 = 3x - 3
\]
3. Solve for \(x\):
\[
4x - 3x = -3 - 1
\]
\[
x = -4
\]

Final Answers:

\[
\boxed{
\begin{array}{ll}
a) & \lambda = 1 \\
b) & \lambda = -4 \\
c) & \lambda = 1 \\
d) & \lambda = 2 \\
e) & x = \frac{1}{2} \\
f) & \lambda = -4 \\
g) & x = 1 \\
h) & x = \frac{15}{2} \\
i) & x = \frac{1}{2} \\
j) & x = 0 \\
k) & x = \frac{3}{7} \\
l) & x = -12 \\
m) & x = -3 \\
n) & \lambda = \frac{-3x + 1}{3} \\
o) & x = 2 \\
p) & x = 2 \\
q) & x = \frac{7}{3} \\
r) & x = \frac{6}{7} \\
s) & x = \frac{4}{3} \\
t) & x = \frac{1}{2} \\
u) & x = \frac{5}{2} \\
v) & x = -\frac{2}{3} \\
w) & x = -\frac{5}{11} \\
x) & x = -2 \\
y) & x = -\frac{2}{3} \\
z) & x = -4 \\
\end{array}
}
\]