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Let's solve each of the exponential equations step by step, without using logarithms.

### 1. \( 4^{x+3} = 1 \)
For any base \(a\), \(a^0 = 1\). Thus:

\[
4^{x+3} = 4^0 \implies x + 3 = 0
\]

Solving for \(x\):

\[
x = -3
\]

### 2. \( 5^{3-2x} = 125 \)
First, express \(125\) as a power of \(5\):

\[
125 = 5^3
\]

Now, equate the exponents:

\[
3 - 2x = 3
\]

\[
-2x = 0 \implies x = 0
\]

### 3. \( 3^{1-2x} = 243 \)
First, express \(243\) as a power of \(3\):

\[
243 = 3^5
\]

Now, equate the exponents:

\[
1 - 2x = 5
\]

\[
-2x = 4 \implies x = -2
\]

### 4. \( 3^{2x} = 3^{-x} \)
Since the bases are the same, set the exponents equal to each other:

\[
2x = -x
\]

\[
3x = 0 \implies x = 0
\]

### 5. \( 8^{2x+5} = 4096 \)
Express \(4096\) as a power of \(8\):

\[
8^4 = 4096
\]

Now, equate the exponents:

\[
2x + 5 = 4
\]

\[
2x = -1 \implies x = -\frac{1}{2}
\]

### 6. \( 4^x = 64 \)
Express \(64\) as a power of \(4\):

\[
64 = 4^{3/2}
\]

Now, equate the exponents:

\[
x = \frac 3

Solve the following exponential equations without using logarithms:
1. 4^(x+3) = 1
2. 5^(3-2x) = 125
3. 3^(1-2x) = 243
4. 3^(2x) = 3^(-x)
5. 8^(2x+5) = 4096
6. 4^x = 64

Learn how to solve a variety of exponential equations, including equations with different bases. This guide provides detailed step-by-step solutions for students in Grades 9-12.

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