Solve Exponential Equations

Note that 27, 9, and 3 may be written as powers of 3 as follows:

$$ 27 = 3^3, \quad 9 = 3^2, \quad \text{and} \quad 3 = 3^1 $$

Using the above and the formula for negative exponents $\frac{1}{x^n} = x^{-n}$, we rewrite the given equation:

$$ (3^3)^{2x} \cdot (3^{-2})^{x - 2} = (3^2)^{-x} \cdot (3^{-1})^{2 - x} $$

We now use the formula $(x^m)^n = x^{mn}$:

$$ 3^{6x} \cdot 3^{-2x + 4} = 3^{-2x} \cdot 3^{-2 + x} $$

Use the formula $x^m \cdot x^n = x^{m + n}$ to combine the exponents on each side:

$$ 3^{6x - 2x + 4} = 3^{-2x - 2 + x} $$

Simplify the exponents to obtain:

$$ 3^{4x + 4} = 3^{-x - 2} $$

We use the fact that an exponential function is a one-to-one function. If $3^m = 3^n$, then $m = n$. We can equate the exponents:

$$ 4x + 4 = -x - 2 $$

Solve the above for $x$:

$$ 5x = -6 \implies x = \frac{-6}{5} = -1.2 $$

Below is shown the graph of the left side of the given equation when written with the right side equal to zero. The $x$-intercept is an approximation to the analytical solution.

Note that $\sqrt{16^x} = (16^x)^{1/2} = 16^{(1/2)x}$ and use it to rewrite the given equation:

$$ 16^{(1/2)x} = \left(\frac{1}{2}\right)^{x^2 - 1} $$

Note that $\frac{1}{2} = 2^{-1}$ and $16 = 2^4$. Use these to rewrite the equation in base 2:

$$ (2^4)^{(1/2)x} = (2^{-1})^{x^2 - 1} $$

Use the formula $(x^m)^n = x^{mn}$ to simplify:

$$ 2^{2x} = 2^{-x^2 + 1} $$

Because the bases are identical, we can equate the exponents:

$$ 2x = -x^2 + 1 $$

Rewrite in standard quadratic form and solve:

$$ x^2 + 2x - 1 = 0 $$

Using the quadratic formula, $\Delta = 2^2 - 4(1)(-1) = 8$. This yields two solutions:

$$ x_1 = -1 - \sqrt{2} \approx -2.41 \quad \text{and} \quad x_2 = -1 + \sqrt{2} \approx 0.41 $$

Let $u = 2^x$. This means $2^{-x} = \frac{1}{2^x} = \frac{1}{u}$. Rewrite the equation in terms of $u$:

$$ u - \frac{6}{u} = 6 $$

$u$ cannot be zero (because $2^x > 0$). Multiply all terms by $u$ and simplify:

$$ u \left( u - \frac{6}{u} \right) = 6u $$ $$ u^2 - 6 = 6u $$ $$ u^2 - 6u - 6 = 0 $$

Solve the quadratic equation for $u$ using the quadratic formula:

$$ u_1 = 3 + \sqrt{15} \quad \text{and} \quad u_2 = 3 - \sqrt{15} $$

We now solve for $x$ using the substitution $u = 2^x$.

First Equation: $2^x = 3 + \sqrt{15}$

$$ \ln(2^x) = \ln(3 + \sqrt{15}) $$ $$ x \ln(2) = \ln(3 + \sqrt{15}) $$ $$ x = \frac{\ln(3 + \sqrt{15})}{\ln 2} \approx 2.78 $$

Second Equation: $2^x = 3 - \sqrt{15}$. This has no solution because $3 - \sqrt{15}$ is a negative number, and $2^x$ must be strictly positive.

Take the natural logarithm ($\ln$) of both sides of the given equation:

$$ \ln(5^{x + 1}) = \ln(100 \cdot 3^x) $$

Use logarithm rules to simplify and bring the exponents down:

$$ (x + 1) \ln 5 = \ln 100 + x \ln 3 $$

Expand the left side and group all terms containing $x$ on one side:

$$ x \ln 5 + \ln 5 = \ln 100 + x \ln 3 $$ $$ x \ln 5 - x \ln 3 = \ln 100 - \ln 5 $$

Factor out $x$ and solve:

$$ x (\ln 5 - \ln 3) = \ln 100 - \ln 5 $$ $$ x = \frac{\ln 100 - \ln 5}{\ln 5 - \ln 3} \approx 5.86 $$

Let $u = e^x$, which gives $e^{-x} = \frac{1}{u}$. Substitute this into the equation:

$$ \frac{u - \frac{1}{u}}{2} = 3 $$

Multiply both sides by 2 and simplify:

$$ u - \frac{1}{u} = 6 $$

Multiply the entire equation by $u$ to clear the fraction:

$$ u^2 - 1 = 6u \implies u^2 - 6u - 1 = 0 $$

Solve for $u$ using the quadratic formula:

$$ u_1 = 3 + \sqrt{10} \quad \text{and} \quad u_2 = 3 - \sqrt{10} $$

Now back-substitute $u = e^x$.

First Equation: $e^x = 3 + \sqrt{10}$

$$ \ln(e^x) = \ln(3 + \sqrt{10}) \implies x = \ln(3 + \sqrt{10}) \approx 1.82 $$

Second Equation: $e^x = 3 - \sqrt{10}$. This has no solution since $3 - \sqrt{10} < 0$.

Take the natural logarithm ($\ln$) of both sides:

$$ \ln(3^{2 - 3x}) = \ln(4^{2x + 1}) $$

Use log rules to bring the exponents to the front:

$$ (2 - 3x)\ln 3 = (2x + 1)\ln 4 $$

Expand both sides of the equation and group the terms with $x$:

$$ 2\ln 3 - 3x\ln 3 = 2x\ln 4 + \ln 4 $$ $$ 2\ln 3 - \ln 4 = 3x\ln 3 + 2x\ln 4 $$ $$ x(3\ln 3 + 2\ln 4) = 2\ln 3 - \ln 4 $$ $$ x = \frac{2\ln 3 - \ln 4}{3\ln 3 + 2\ln 4} \approx 0.13 $$

Note that $9^x = (3^2)^x = 3^{2x} = (3^x)^2$. Rewrite the given equation using this and expand $3^{x+1}$ as $3 \cdot 3^x$:

$$ (3^x)^2 - 3 \cdot 3^x + 2 = 0 $$

Let $u = 3^x$ and rewrite the equation as a standard quadratic in $u$:

$$ u^2 - 3u + 2 = 0 $$

Solve for $u$ by factoring:

$$ (u - 2)(u - 1) = 0 $$

This gives two solutions for $u$: $u_1 = 2$ and $u_2 = 1$. Now substitute $3^x$ back in.

First Equation: $3^x = 2$

$$ \ln(3^x) = \ln 2 \implies x \ln 3 = \ln 2 \implies x = \frac{\ln 2}{\ln 3} \approx 0.63 $$

Second Equation: $3^x = 1$. The only exponent that yields 1 is 0, so $x = 0$.

1. Standardize the Base:
Express $4^x$ as $(2^2)^x = (2^x)^2$. Expand $2^{x+2}$ using exponent rules: $2^{x+2} = 2^x \cdot 2^2 = 4 \cdot 2^x$.

$$ (2^x)^2 - 4(2^x) - 32 = 0 $$

2. Substitute:
Let $u = 2^x$. The equation becomes a straightforward quadratic:

$$ u^2 - 4u - 32 = 0 $$

3. Factor and Solve for $u$:
We need two numbers that multiply to $-32$ and add to $-4$. These are $-8$ and $+4$.

$$ (u - 8)(u + 4) = 0 $$

This gives us $u = 8$ or $u = -4$.

4. Solve for $x$:
Substitute $2^x$ back in for $u$.

• Case 1: $2^x = 8 \implies 2^x = 2^3 \implies \mathbf{x = 3}$
• Case 2: $2^x = -4$. This has no solution since $2^x$ must be strictly positive.

The only solution is $\mathbf{x = 3}$.

For an equation of the form $A^B = 1$, there are three distinct mathematical conditions under which this is true. We will use a table to systematically check all three cases.

Condition for $A^B = 1$	Application to $x^{x^2 - 5x + 6} = 1$
Case 1: The Base is 1 ($A = 1$, $B$ can be anything)	Set the base equal to 1: $x = 1$. (Check: $1^{(1-5+6)} = 1^2 = 1$. Valid.) Solution: $x = 1$
Case 2: The Exponent is 0 ($B = 0$, $A \neq 0$)	Set the exponent equal to 0: $x^2 - 5x + 6 = 0$ $(x - 2)(x - 3) = 0 \implies x = 2, x = 3$. (Check bases: Neither base is 0. Valid.) Solutions: $x = 2, x = 3$
Case 3: The Base is -1 ($A = -1$, and $B$ must be an even integer)	Set the base equal to -1: $x = -1$. Now verify if the exponent is even when $x = -1$: $B = (-1)^2 - 5(-1) + 6 = 1 + 5 + 6 = 12$. Since 12 is even, $(-1)^{12} = 1$. Valid. Solution: $x = -1$

Final Solution Set:
The complete set of real solutions is $\mathbf{x \in \{-1, 1, 2, 3\}}$.

1. Combine the Left Side:
Since both $2$ and $5$ are raised to the exact same exponent, we can use the rule $a^n \cdot b^n = (a \cdot b)^n$ to combine them:

$$ (2 \cdot 5)^{3x} = 10000^{x-1} $$ $$ 10^{3x} = 10000^{x-1} $$

2. Create a Common Base:
Recognize that $10000$ is a power of 10. Specifically, $10000 = 10^4$. Substitute this into the right side:

$$ 10^{3x} = (10^4)^{x-1} $$

3. Distribute the Exponent:
Use the power rule $(x^m)^n = x^{mn}$:

$$ 10^{3x} = 10^{4(x-1)} $$ $$ 10^{3x} = 10^{4x - 4} $$

4. Equate and Solve:
Because the bases are equal, the exponents must be equal:

$$ 3x = 4x - 4 $$ $$ 3x - 4x = -4 $$ $$ -x = -4 \implies \mathbf{x = 4} $$