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 \( x \) 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.

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 become a useful tool. By applying logarithms, we can solve for \(x\) without needing to find complex decimal values.

To illustrate the process, consider the equation \(10^{x + 64} = 100\). The first step is to isolate the exponential expression. By subtracting 64 from both sides, we rewrite the equation as \(10^x = 36\). 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 down the exponent \(x\):

\(x \cdot \log(10) = \log(36)\)

Since \(\log(10) = 1\), this simplifies to:

\(x = \log(36)\

This expression is a valid answer, but if an approximation is required, you can use a calculator to find the numerical value.

In another example, consider the equation \(3 = 2^{x + 1}\). Here, the exponential expression is already isolated, so we proceed to take the natural logarithm (since the base is not 10) of both sides:

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

Applying the power rule again, we have:

\(\ln(3) = (x + 1) \cdot \ln(2)\)

To isolate \(x\), we divide both sides by \(\ln(2)\):

\(x + 1 = \frac{\ln(3)}{\ln(2)}\)

Finally, subtracting 1 gives us:

\(x = \frac{\ln(3)}{\ln(2)} - 1\

This expression can also be approximated using a calculator, yielding a numerical value of approximately \(0.58\).

By mastering the use of logarithms and understanding how to manipulate exponential equations, you can confidently tackle a wide range of problems involving exponential relationships.

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. In this case, you can set the arguments of the logarithms equal to each other, transforming the problem into a basic linear equation. For example, if you have log2(x + 5) = log2(9), you can directly set x + 5 = 9 and solve for x, yielding x = 4.

The second type of logarithmic equation features a single logarithm set equal to a constant. To solve this, you convert the logarithmic expression into its exponential form. For instance, if you have log2(4x) = 5, you start by rewriting it in exponential form: 25 = 4x. This simplifies to 32 = 4x. Dividing both sides by 4 gives x = 8.

It’s crucial to check your solutions, especially when dealing with logarithms. After finding x = 8, substitute it back into the argument of the logarithm, which is 4x. In this case, 4 * 8 = 32, a positive number, confirming that the solution is valid. Remember, logarithms are undefined for non-positive values, so if your substitution yields a negative result, that solution is not acceptable.

By mastering these two types of logarithmic equations and understanding how to manipulate them into linear forms or exponential forms, you can confidently tackle a variety of problems. Practice is key to becoming proficient in solving logarithmic equations, so keep working through examples to reinforce your understanding.