Problem 19 - Precalculus : Building Concepts and Connections

Question

Solve the exponential equation. Round to three decimal places, when needed. $$3^{-2 x-1}=2^{x}$$

Step-by-Step Solution

Answer

The solution to the equation \(3^{-2x -1} = 2^{x}\) is approximately \(x =\frac{-log(3)}{2log(3) + log(2)}\)

Step 1: Rewrite equation

Rewrite the equation \(3^{-2x -1} = 2^{x}\) as \(\frac{1} {3^{2x + 1}} = 2^{x}\) to make it easier to solve.

Step 2: Apply Logarithmic Function

Using logarithm to base 10, apply log to both sides of the equation: \(log(\frac{1} {3^{2x + 1}}) = log(2^{x}) \) yields \( -(2x+1)log(3) = xlog(2) \)

Step 3: Simplify and Solve for x

Expand and rearrange the terms gives \(2xlog(3) + log(3) + xlog(2) = 0\), which can be written as \(x(2log(3) + log(2)) = -log(3) \). Finally we divide both sides by \(2log(3) + log(2)\) to isolate x, that gives us \(x =\frac {-log(3)} {2log(3) + log(2)}\).

Key Concepts

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions and are exceptionally useful in solving exponential equations. The logarithm of a number essentially asks the question, 'To what power should the base be raised to obtain that number?' For example, the expression \(\log_b(A)\) is read as 'the logarithm of A with base b' and it means “What power should we raise b to get A?”

Logarithms have a few key components: the base (b), the argument (A), and the resulting exponent. A standard logarithmic equation is \(b^y = A\), which can be represented in logarithmic form as \(y = \log_b(A)\). This property makes logarithms particularly valuable for turning multiplicative relationships into additive ones, which simplifies solving equations extensively. Most commonly, you will see the natural logarithm (base \(e\), where \(e\) is Euler's number) or the common logarithm (base 10) being used.

Exponential Equation Step by Step

When tackling an exponential equation step by step, we start by identifying and isolating the exponential part of the equation. For the given exercise \(3^{-2x-1} = 2^x\), the exponential equation involves different bases, which can add a layer of complexity.

The key to solving such equations often lies in using logarithms because they allow us to bring the exponents down to the level of coefficients. Here's how you might approach it:

• Rewrite the equation to make the exponential parts clear and isolated if possible.
• Apply a logarithmic function to both sides of the equation to remove the exponent. The base of the logarithm can be any number, but common bases like 10 or \(e\) can make calculations easier.
• Use the properties of logarithms to simplify the equation further if possible.
• Solve the resulting expression for the variable of interest, in this case, x.

Each of these steps is a move towards transforming the equation into a form where the variable can be isolated and solved for, using basic algebraic operations.

Properties of Logarithms

Mastering the properties of logarithms is essential for solving logarithmic and exponential equations. Some of the key properties include:

• Product Rule: The logarithm of a product is equal to the sum of the logarithms of the factors (\(\log_b(MN) = \log_b(M) + \log_b(N)\)).
• Quotient Rule: The logarithm of a quotient is the difference of the logarithms (\(\log_b(\frac{M}{N}) = \log_b(M) - \log_b(N)\)).
• Power Rule: The logarithm of an exponentiated number allows the exponent to be pulled out as a coefficient (\(\log_b(M^p) = p\log_b(M)\)).
• Change of Base Formula: To change the base of a logarithmic expression, you can use (\(\log_b(M) = \frac{\log_k(M)}{\log_k(b)}\)), where \(k\) is a new base and \(b\) is the old base.

Understanding and applying these properties can simplify complex logarithmic expressions and can turn the process of solving an exponential equation into a manageable task. In the provided exercise, both the quotient and the power rules were applied to simplify the equation and bring x down out of the exponent, this enabled us to solve for x algebraically.