Exponential & 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 \( x \) 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 rewrite \( 16 \) as a power of \( 2 \). Since \( 16 = 2^4 \), we can express the equation as \( 2^4 = 2^x \). With both sides having the same base, we can set the exponents equal to each other, resulting in \( 4 = x \). Thus, the solution is \( x = 4 \).

In another example, we have \( 64 = 2^x \). To rewrite \( 64 \) in terms of base \( 2 \), we recognize that \( 64 = 8^2 \) and \( 8 = 2^3 \). Therefore, \( 64 = (2^3)^2 = 2^6 \). This gives us \( 2^6 = 2^x \), leading to \( 6 = x \) and the solution \( x = 6 \).

Next, consider the equation \( 5^{x+1} = \sqrt{5} \). Here, both sides already have the base \( 5 \). We can rewrite \( \sqrt{5} \) as \( 5^{1/2} \), resulting in \( 5^{x+1} = 5^{1/2} \). Setting the exponents equal gives us the equation \( x + 1 = \frac{1}{2} \). Solving for \( x \) by subtracting \( 1 \) from both sides yields \( x = \frac{1}{2} - 1 = -\frac{1}{2} \).

Lastly, consider \( 27 = 9^x \). Since \( 9 \) cannot be rewritten as a power of \( 27 \), we need to express both sides with a common base. We know \( 9 = 3^2 \), so we rewrite the equation as \( 27 = (3^2)^x = 3^{2x} \). Recognizing that \( 27 = 3^3 \), we can set the exponents equal: \( 3 = 2x \). Dividing both sides by \( 2 \) gives us \( x = \frac{3}{2} \).

By rewriting exponential equations to have the same base, we can transform them into linear equations, making it easier to solve for \( x \). This method is a powerful tool for tackling a variety of exponential equations.

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, logarithms provide a powerful alternative. By applying logarithmic properties, we can effectively isolate the variable in the exponent.

To illustrate this process, consider the equation \( 10^{x + 64} = 100 \). The first step is to isolate the exponential expression. By subtracting 64 from both sides, we simplify the equation to \( 10^x = 36 \). Next, we determine which logarithm to use. Since the base of the exponential expression is 10, we apply the common logarithm (logarithm base 10) to both sides, resulting in \( \log(10^x) = \log(36) \).

Utilizing the power rule of logarithms, we can bring the exponent down: \( x \cdot \log(10) = \log(36) \). Knowing that \( \log(10) = 1 \), we simplify this to \( x = \log(36) \). This expression is valid as is, but if an approximation is required, we can calculate \( \log(36) \) using a calculator, yielding approximately \( x \approx 1.56 \).

In another example, consider the equation \( 3 = 2^{x + 1} \). Here, the exponential expression is already isolated, so we proceed directly to taking the natural logarithm (since the base is not 10). This gives us \( \ln(3) = \ln(2^{x + 1}) \). Again, applying the power rule allows us to rewrite this as \( \ln(3) = (x + 1) \cdot \ln(2) \).

To isolate \( x \), we divide both sides by \( \ln(2) \), resulting in \( \frac{\ln(3)}{\ln(2)} = x + 1 \). Finally, we subtract 1 from both sides to find \( x = \frac{\ln(3)}{\ln(2)} - 1 \). This expression can also be approximated using a calculator, yielding \( x \approx 0.58 \).

By mastering the use of logarithms and understanding how to manipulate exponential equations, you can confidently tackle a wide range of problems involving exponents. Whether you are working with common or natural logarithms, the key steps involve isolating the exponential term, applying the appropriate logarithm, and using logarithmic properties to solve for the variable.

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 + 1) = log2(5), you can simplify this to x + 1 = 5. 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 (base e), 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. Calculating 25 gives 32, leading to the equation 32 = 4x. Dividing both sides by 4 results in x = 8.

It is crucial to check your solution by substituting back into the original logarithmic expression. For instance, substituting x = 8 into 4x gives 4(8) = 32, which is positive. Since logarithms are only defined for positive arguments, this confirms that x = 8 is a valid solution.

In summary, solving logarithmic equations involves recognizing the type of equation you are dealing with, applying logarithmic properties, converting to exponential form when necessary, and ensuring that the solutions yield positive arguments for the logarithms. With practice, these concepts will become second nature.