Question

Solve each exponential equation. Express irrational solutions in exact form. $$ 2^{x+1}=5^{1-2 x} $$

Short answer

\( x = \frac{ \ln(5) - \ln(2)}{\ln(2) + 2\ln(5)} \)

Step-by-step solution

Take the logarithm of both sides

Apply the natural logarithm (ln) to both sides of the equation. \[ \ln(2^{x+1}) = \ln(5^{1-2x}) \]

Apply the power rule of logarithms

Use the power rule of logarithms, which states that \( \ln(a^b) = b \cdot \ln(a) \). \[ (x+1) \ln(2) = (1-2x) \ln(5) \]

Distribute the logarithm

Distribute the logarithms across the terms in the parentheses. \[ x \ln(2) + \ln(2) = \ln(5) - 2x \ln(5) \]

Isolate the variable terms

Combine the \( x \)-terms on one side of the equation. \[ x \ln(2) + 2x \ln(5) = \ln(5) - \ln(2) \]

Factor out x

Factor out \( x \) from the left-hand side. \[ x ( \ln(2) + 2 \ln(5) ) = \ln(5) - \ln(2) \]

Solve for x

Divide both sides by the coefficient \( \ln(2) + 2 \ln(5) \). \[ x = \frac{ \ln(5) - \ln(2)}{\ln(2) + 2 \ln(5)} \]

Key concepts

logarithms

Logarithms are the inverse operations of exponentiation. If you have an exponential equation like \(a^b = c\), the logarithm helps you solve for one of the variables. For base \(a\), we write this as \(\text{log}_a(c) = b\). In simpler terms, logarithms tell you the power to which you must raise the base to get a particular number.

For example, \( \text{log}_2(8) = 3 \) because \( 2^3 = 8\). Logarithms allow us to transform problems involving multiplication into simpler addition problems.
Different bases are used in logarithms, such as base 10 (common logarithms) and the natural base \(e\) (natural logarithms).

When dealing with exponential equations, taking the logarithm of both sides can make it easier to solve for the unknown variable.

natural logarithm

The natural logarithm, often denoted as \( \text{ln}\), uses the constant \( e \) (approximately 2.71828) as the base. It's prevalent in calculus because of its unique properties with respect to differentiation and integration.
For instance, the derivative of \( \text{ln}(x) \) with respect to \( x \) is \( \frac{1}{x} \).

In the exercise solution, we use the natural logarithm to simplify the exponential equation and utilize its properties, particularly when applying the power rule. This helps break down complex exponentiation into manageable linear terms.

power rule

The power rule for logarithms states that \( \text{ln}(a^b) = b \text{ln}(a) \).
This means that you can move the exponent in an exponential expression to the front as a coefficient.

The power rule is extremely useful when solving exponential equations because it allows us to linearize the equation.
For example, if we have \( \text{ln}(2^{x+1}) = \text{ln}(5^{1-2x}) \), applying the power rule transforms it to \( (x+1) \text{ln}(2) = (1-2x) \text{ln}(5) \).
This simplification is crucial as it sets up the stage for solving the variable by rearranging and factoring terms. Without the power rule, working with exponential equations would be much more complex.

By applying the power rule in our exercise, we could manipulate the logarithmic terms to isolate \( x \) and eventually find its exact value.