When working with exponential functions, evaluating the function for a specific value of \( x \) is straightforward. However, when the function is set equal to a value, the goal shifts to finding the \( x \) that satisfies the equation. Fortunately, this process can be simplified by rewriting both sides of the equation to have the same base, allowing us to solve it as a basic linear equation.
For example, consider the equation \( 16 = 2^x \). To solve for \( x \), we first express both sides with the same base. We know that \( 16 \) can be rewritten as \( 2^4 \). Thus, we have:
\[ 2^4 = 2^x \]
Since the bases are the same, we can set the exponents equal to each other:
\[ 4 = x \]
This gives us the solution \( x = 4 \).
In another example, let’s solve \( 64 = 2^x \). We need to express \( 64 \) as a power of \( 2 \). We can break it down as follows:
\[ 64 = 8^2 = (2^3)^2 = 2^6 \]
Now we rewrite the equation:
\[ 2^6 = 2^x \]
Setting the exponents equal gives us:
\[ 6 = x \]
Next, consider the equation \( 5^{x+1} = \sqrt{5} \). The square root can be expressed as an exponent:
\[ \sqrt{5} = 5^{1/2} \]
Now we have:
\[ 5^{x+1} = 5^{1/2} \]
Setting the exponents equal results in:
\[ x + 1 = \frac{1}{2} \]
Solving for \( x \) gives:
\[ x = \frac{1}{2} - 1 = -\frac{1}{2} \]
Lastly, let’s solve \( 27 = 9^x \). Since \( 9 \) can be rewritten as \( 3^2 \), we have:
\[ 9^x = (3^2)^x = 3^{2x} \]
Now, we rewrite \( 27 \) as \( 3^3 \), leading to:
\[ 3^3 = 3^{2x} \]
Setting the exponents equal gives:
\[ 3 = 2x \]
Dividing both sides by \( 2 \) results in:
\[ x = \frac{3}{2} \]
By rewriting exponential equations to have the same base, we can easily solve for \( x \) using basic algebraic techniques. This method is effective for a variety of exponential equations, allowing for straightforward solutions.
15 Terms
