kuta software - infinite algebra 2
exponential equations not requiring logarithms
solve each equation.
1) $4^{2x + 3} = 1$
2) $5^{3 - 2x} = 5^{-x}$
3) $3^{1 - 2x} = 243$
4) $3^{2a} = 3^{-a}$
5) $4^{3x - 2} = 1$
6) $4^{2p} = 4^{-2p - 1}$
7) $6^{-2a} = 6^{2 - 3a}$
8) $2^{2x + 2} = 2^{3x}$
9) $6^{3m} \cdot 6^{-m} = 6^{-2m}$
10) $\frac{2^x}{2^x} = 2^{-2x}$
11) $10^{-3x} \cdot 10^x = \frac{1}{10}$
12) $3^{-2x + 1} \cdot 3^{-2x - 3} = 3^{-x}$

Question

Accepted Answer

### Problem 1: $ 4^{2x + 3} = 1 $
# Explanation:
## Step 1: Recall that $ a^0 = 1 $ for $ a
eq 0 $
So we can rewrite the right - hand side as $ 4^0 $. Then the equation becomes $ 4^{2x+3}=4^0 $.
## Step 2: Use the property of exponential functions: if $ a^m=a^n $, then $ m = n $ (for $ a>0,a
eq1 $)
Since the base $ 4>0 $ and $ 4
eq1 $, we can set the exponents equal to each other: $ 2x + 3=0 $.
## Step 3: Solve for $ x $
Subtract 3 from both sides: $ 2x=-3 $. Then divide both sides by 2: $ x =-\frac{3}{2} $.
# Answer: $ x =-\frac{3}{2} $

### Problem 2: $ 5^{3 - 2x}=5^{-x} $
# Explanation:
## Step 1: Use the property of exponential functions $ a^m=a^n\Rightarrow m = n $ ( $ a = 5>0,a
eq1 $)
Set the exponents equal: $ 3-2x=-x $.
## Step 2: Solve for $ x $
Add $ 2x $ to both sides: $ 3=x $.
# Answer: $ x = 3 $

### Problem 3: $ 3^{1 - 2x}=243 $
# Explanation:
## Step 1: Rewrite 243 as a power of 3
We know that $ 3^5 = 243 $, so the equation becomes $ 3^{1-2x}=3^5 $.
## Step 2: Use the property $ a^m=a^n\Rightarrow m = n $ ( $ a = 3>0,a
eq1 $)
Set the exponents equal: $ 1-2x = 5 $.
## Step 3: Solve for $ x $
Subtract 1 from both sides: $ - 2x=4 $. Divide both sides by - 2: $ x=-2 $.
# Answer: $ x=-2 $

### Problem 4: $ 3^{2a}=3^{-a} $
# Explanation:
## Step 1: Use the property $ a^m=a^n\Rightarrow m = n $ ( $ a = 3>0,a
eq1 $)
Set the exponents equal: $ 2a=-a $.
## Step 2: Solve for $ a $
Add $ a $ to both sides: $ 3a = 0 $. Divide both sides by 3: $ a = 0 $.
# Answer: $ a = 0 $

### Problem 5: $ 4^{3x - 2}=1 $
# Explanation:
## Step 1: Rewrite 1 as $ 4^0 $
The equation becomes $ 4^{3x - 2}=4^0 $.
## Step 2: Use the property $ a^m=a^n\Rightarrow m = n $ ( $ a = 4>0,a
eq1 $)
Set the exponents equal: $ 3x-2 = 0 $.
## Step 3: Solve for $ x $
Add 2 to both sides: $ 3x=2 $. Divide both sides by 3: $ x=\frac{2}{3} $.
# Answer: $ x=\frac{2}{3} $

### Problem 6: $ 4^{2p}=4^{-2p - 1} $
# Explanation:
## Step 1: Use the property $ a^m=a^n\Rightarrow m = n $ ( $ a = 4>0,a
eq1 $)
Set the exponents equal: $ 2p=-2p - 1 $.
## Step 2: Solve for $ p $
Add $ 2p $ to both sides: $ 4p=-1 $. Divide both sides by 4: $ p=-\frac{1}{4} $.
# Answer: $ p =-\frac{1}{4} $

### Problem 7: $ 6^{-2a}=6^{2 - 3a} $
# Explanation:
## Step 1: Use the property $ a^m=a^n\Rightarrow m = n $ ( $ a = 6>0,a
eq1 $)
Set the exponents equal: $ -2a=2 - 3a $.
## Step 2: Solve for $ a $
Add $ 3a $ to both sides: $ a = 2 $.
# Answer: $ a = 2 $

### Problem 8: $ 2^{2x + 2}=2^{3x} $
# Explanation:
## Step 1: Use the property $ a^m=a^n\Rightarrow m = n $ ( $ a = 2>0,a
eq1 $)
Set the exponents equal: $ 2x + 2=3x $.
## Step 2: Solve for $ x $
Subtract $ 2x $ from both sides: $ x = 2 $.
# Answer: $ x = 2 $

### Problem 9: $ 6^{3m}\cdot6^{-m}=6^{-2m} $
# Explanation:
## Step 1: Use the property of exponents $ a^m\cdot a^n=a^{m + n} $
Left - hand side: $ 6^{3m+( - m)}=6^{2m} $. So the equation becomes $ 6^{2m}=6^{-2m} $.
## Step 2: Use the property $ a^m=a^n\Rightarrow m = n $ ( $ a = 6>0,a
eq1 $)
Set the exponents equal: $ 2m=-2m $.
## Step 3: Solve for $ m $
Add $ 2m $ to both sides: $ 4m = 0 $. Divide both sides by 4: $ m = 0 $.
# Answer: $ m = 0 $

### Problem 10: $ \frac{2^x}{2^x}=2^{-2x} $
# Explanation:
## Step 1: Simplify the left - hand side
$ \frac{2^x}{2^x}=1 $, and $ 1 = 2^0 $, so the equation becomes $ 2^0=2^{-2x} $.
## Step 2: Use the property $ a^m=a^n\Rightarrow m = n $ ( $ a = 2>0,a
eq1 $)
Set the exponents equal: $ 0=-2x $.
## Step 3: Solve for $ x $
Divide both sides by - 2: $ x = 0 $.
# Answer: $ x = 0 $

### Problem 11: $ 10^{-3x}\cdot10^{x}=\frac{1}{10} $
# Explanation:
## Step 1: Use the property $ a^m\cdot a^n=a^{m + n} $ and rewrite $ \frac{1}{10} $ as $ 10^{-1} $
Left - hand side: $ 10^{-3x + x}=10^{-2x} $. The equation becomes $ 10^{-2x}=10^{-1} $.
## Step 2: Use the property $ a^m=a^n\Rightarrow m = n $ ( $ a = 10>0,a
eq1 $)
Set the exponents equal: $ -2x=-1 $.
## Step 3: Solve for $ x $
Divide both sides by - 2: $ x=\frac{1}{2} $.
# Answer: $ x=\frac{1}{2} $

### Problem 12: $ 3^{-2x + 1}\cdot3^{-2x - 3}=3^{-x} $
# Explanation:
## Step 1: Use the property $ a^m\cdot a^n=a^{m + n} $
Left - hand side: $ 3^{(-2x + 1)+(-2x - 3)}=3^{-4x-2} $. So the equation becomes $ 3^{-4x - 2}=3^{-x} $.
## Step 2: Use the property $ a^m=a^n\Rightarrow m = n $ ( $ a = 3>0,a
eq1 $)
Set the exponents equal: $ -4x-2=-x $.
## Step 3: Solve for $ x $
Add $ 4x $ to both sides: $ -2 = 3x $. Divide both sides by 3: $ x=-\frac{2}{3} $.
# Answer: $ x =-\frac{2}{3} $

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QUESTION IMAGE

Question

Explanation:

Problem 1: \( 4^{2x + 3} = 1 \)

Step 1: Recall that \( a^0 = 1 \) for \( a

Step 2: Use the property of exponential functions: if \( a^m=a^n \), then \( m = n \) (for \( a>0,a

Step 3: Solve for \( x \)

Step 1: Use the property of exponential functions \( a^m=a^n\Rightarrow m = n \) ( \( a = 5>0,a

Step 2: Solve for \( x \)

Step 1: Rewrite 243 as a power of 3

Step 2: Use the property \( a^m=a^n\Rightarrow m = n \) ( \( a = 3>0,a

Step 3: Solve for \( x \)

Answer:

Problem 2: \( 5^{3 - 2x}=5^{-x} \)