Solving Exponential and Logarithmic Equations: Videos & Practice Problems

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 value of \( x \) 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, the equation becomes \( 2^4 = 2^x \). Since the bases are the same, we can set the exponents equal to each other, leading to \( 4 = x \). Therefore, the solution is \( x = 4 \).

In another example, we have \( 64 = 2^x \). To rewrite \( 64 \) in terms of base \( 2 \), we can express it as \( 8^2 \) and recognize that \( 8 = 2^3 \). Thus, \( 64 = (2^3)^2 = 2^6 \). Now, we have \( 2^6 = 2^x \), allowing us to set the exponents equal: \( 6 = x \). Hence, \( x = 6 \).

Next, consider the equation \( 5^{x+1} = \sqrt{5} \). The square root can be rewritten as an exponent: \( \sqrt{5} = 5^{1/2} \). This gives us \( 5^{x+1} = 5^{1/2} \). With the same base, we equate the exponents: \( x + 1 = \frac{1}{2} \). Solving for \( x \) yields \( x = \frac{1}{2} - 1 = -\frac{1}{2} \).

Lastly, consider \( 27 = 9^x \). Since \( 9 \) can be expressed as \( 3^2 \), we rewrite the equation as \( 27 = (3^2)^x = 3^{2x} \). Knowing that \( 27 = 3^3 \), we can set the bases equal: \( 3^3 = 3^{2x} \). This leads to \( 3 = 2x \). Dividing both sides by \( 2 \) gives \( x = \frac{3}{2} \).

In summary, solving exponential equations often involves rewriting the expressions to have the same base, allowing for straightforward comparison of exponents. This method simplifies the process and reinforces the understanding of exponential relationships.

Exponential equations can often be solved by rewriting both sides to have the same base, allowing us to set the exponents equal to each other. However, when faced with an equation like \( 17 = 2^x \), where it’s not straightforward to express 17 as a power of 2, we can utilize logarithms to find the solution. This method is particularly useful when the base of the exponential expression is not easily manipulated.

To solve an exponential equation, follow these steps:

First, isolate the exponential expression. For example, in the equation \( 10^{x} + 64 = 100 \), subtract 64 from both sides to get \( 10^{x} = 36 \). This step is crucial as it sets the stage for applying logarithmic properties.

Next, determine which logarithm to use. If the base of the exponential is 10, use the common logarithm (log). If the base is not 10, use the natural logarithm (ln). In our example, since we have \( 10^{x} \), we take the common log of both sides: \( \log(10^{x}) = \log(36) \).

Then, apply the power rule of logarithms, which states that \( \log(a^b) = b \cdot \log(a) \). This allows us to bring the exponent down: \( x \cdot \log(10) = \log(36) \). Since \( \log(10) = 1 \), this simplifies to \( x = \log(36) \). This result is acceptable as is unless a numerical approximation is required.

For approximation, you can use a calculator to find \( x \approx 1.556 \) since \( \log(36) \) is a constant.

In another example, consider the equation \( 3 = 2^{x+1} \). Start by isolating the exponential expression, which is already done here. Next, since the base is not 10, take the natural log of both sides: \( \ln(3) = \ln(2^{x+1}) \). Again, apply the power rule to get \( \ln(3) = (x+1) \cdot \ln(2) \).

To solve for \( x \), divide both sides by \( \ln(2) \): \( \frac{\ln(3)}{\ln(2)} = x + 1 \). Finally, isolate \( x \) by subtracting 1: \( x = \frac{\ln(3)}{\ln(2)} - 1 \). This expression can also be approximated using a calculator, yielding \( x \approx 0.585 \).

By mastering these steps, you can confidently tackle any exponential equation, whether it involves common or natural logarithms, ensuring you can find solutions even when direct manipulation of the equation is not possible.

Logarithmic equations can be simplified into two main types, making them manageable to solve. The first type involves two logarithms of the same base set equal to each other, while the second type consists of a single logarithm set equal to a constant. Both types ultimately reduce to solving basic linear equations.

For the first type, when you have two logarithms with the same base, you can set the arguments of the logarithms equal to each other. For example, if you have log2(x + 5) = log2(9), you can simplify this to x + 5 = 9. Solving this linear equation gives x = 4.

In a more complex scenario, such as ln(x + 4) - ln(2) = ln(8), you can apply the properties of logarithms to condense the equation. Using the quotient rule, this can be rewritten as ln((x + 4)/2) = ln(8). Since both sides have the same base (natural logarithm), you can set the arguments equal: (x + 4)/2 = 8. Multiplying both sides by 2 results in x + 4 = 16, and isolating x gives x = 12.

The second type of logarithmic equation involves a single logarithm set equal to a constant, such as log2(4x) = 5. To solve this, first express the logarithm in exponential form. This means rewriting the equation as 25 = 4x, which simplifies to 32 = 4x. Dividing both sides by 4 yields x = 8.

It is crucial to verify your solution by substituting it back into the original logarithmic expression. For instance, substituting x = 8 into 4x gives 4 * 8 = 32, which is positive. Since logarithms cannot be taken of negative numbers or zero, this confirms that x = 8 is indeed a valid solution.

In summary, solving logarithmic equations involves recognizing the type of equation you are dealing with, applying logarithmic properties, and ensuring that your solutions yield valid arguments for the logarithmic functions. With practice, these concepts will become second nature.