Solving Exponential Equations Using Bases and Logarithms 🔢

This animation solves the exponential equation $\mathbf{3^{2x-1} = 27}$ by first converting both sides to the same base, which simplifies the problem to a basic linear equation. An alternative solution using the power rule of logarithms is also demonstrated.

Method 1: Equating the Bases (Primary Method)

This method is the most direct when the numbers involved are powers of the same base.

*Step 1: Express Both Sides with the Same Base*

Identify the common base. Since $27$ is a power of $3$ ($27 = 3^3$), we rewrite the equation:
$$
3^{2x-1} = 3^3
$$


*Step 2: Apply the Property of Equal Bases*

The property states that if $a^m = a^n$, then the exponents must be equal ($m=n$). We can drop the bases and equate the exponents:
$$
2x - 1 = 3
$$

*Step 3: Solve the Linear Equation*

This results in a simple two-step algebraic solution:

1. *Add 1 to both sides:*
$$
2x - 1 + 1 = 3 + 1 \implies 2x = 4
$$
2. *Divide by 2:*
$$
\frac{2x}{2} = \frac{4}{2} \implies \mathbf{x = 2}
$$

*Step 4: Verify the Solution*
Substitute $x=2$ back into the original equation:
$$
3^{2(2)-1} = 3^{4-1} = 3^3 = 27 \checkmark
$$

***

Method 2: Using Logarithms (Alternative Method)

This method works for any exponential equation and provides a powerful general technique.

*Step A: Take the Logarithm of Both Sides*

Apply the common logarithm ($\log$) or natural logarithm ($\ln$) to both sides:
$$
\log(3^{2x-1}) = \log(27)
$$

*Step B: Apply the Logarithm Power Rule*

The Power Rule states that $\log(a^m) = m \log(a)$. This allows us to bring the exponent down:
$$
(2x - 1)\log(3) = \log(27)
$$


*Step C: Simplify and Solve for $\mathbf{x}$*

Recognize that $27 = 3^3$, so $\log(27) = \log(3^3) = 3\log(3)$.

$$
(2x - 1)\log(3) = 3\log(3)
$$
Divide both sides by $\log(3)$:
$$
2x - 1 = 3
$$
This reduces the equation back to the same linear equation found in Method 1, yielding the same result:
$$
\mathbf{x = 2}
$$



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