Problem 121 - Elementary and Intermediate Algebra

Question

Explain how to solve the equation \(2^{x+1}=31\)

Step-by-Step Solution

Answer

\(x \approx 3.96\)

Step 1: Isolate the Exponent

Start with the equation \(2^{x+1} = 31\). We have a base of 2 and we need to isolate the expression \(x+1\) in the exponent to solve for \(x\). To do this, recognize that solving \(2^{x+1} = 31\) involves finding \(x+1\) in an exponent of base 2 that equates to a number close to 31.

Step 2: Use Logarithms

Take the logarithm of both sides of the equation to bring down the exponent: \[\log(2^{x+1}) = \log(31)\]Using the logarithm power rule, this becomes:\[(x+1) \cdot \log(2) = \log(31)\]

Step 3: Solve for \(x+1\)

Divide both sides by \(\log(2)\) to solve for \(x+1\):\[x+1 = \frac{\log(31)}{\log(2)}\]Calculate the values:\[\log(31) \approx 1.49136\] and \[\log(2) \approx 0.30103\]Thus, \[x+1 \approx \frac{1.49136}{0.30103} \approx 4.95544\]

Step 4: Isolate \(x\)

Subtract 1 from both sides to solve for \(x\):\[x = 4.95544 - 1\]\[x \approx 3.95544\]

Step 5: Round the Solution

If you need a whole number or a simple decimal approximation, round the solution to 2 decimal places. Thus,\[x \approx 3.96\]

Key Concepts

Logarithms

Logarithms are the inverse operations of exponentiation. Just like subtraction is the inverse operation of addition, logarithms help us find the exponent in exponential equations. For any positive number, the logarithm tells us what power we need to raise another number, called the base, to get that number. This can be written with

• \(b^y = x\) as the exponential form, and
• \(y = \log_b(x)\) as the logarithmic form.

Logarithms are particularly useful when dealing with exponential equations like \(2^{x+1} = 31\), as they allow us to 'bring down' the exponent, making it easier to solve for unknown variables. By applying logarithms to both sides of an equation, the exponent can be expressed as a multiple in a more manageable linear equation. This is done because the logarithm of a power, seen through the Power Rule, transforms multiplication in the logarithmic argument into a multiplication of the logarithm itself. This simplifies the solving process significantly.

Power Rule in Logarithms

The Power Rule is an important tool when solving equations involving exponents and logarithms. This rule states that a logarithm of a power can be simplified by bringing the exponent in the logarithmic expression down in front of the log function. The Power Rule is expressed as:

• \(\log_b(a^c) = c \cdot \log_b(a)\)

In the context of solving \(2^{x+1} = 31\), the Power Rule allows us to rewrite \(\log(2^{x+1})\) as \((x+1) \cdot \log(2)\). This method simplifies the original exponential equation into a straightforward linear equation. Once in this form, finding the solution for \(x\) becomes a matter of simple algebraic manipulation, allowing us to divide both sides by \(\log(2)\) and follow through to solve for the unknown variable.

Rounding Solutions

Rounding is a mathematical process that simplifies numbers, making them easier to work with or communicate. It can be necessary when dealing with long or complex decimal results from calculations, as seen in many log-based solutions of exponential equations. For the equation \(2^{x+1} = 31\), calculating \(x\) yields an approximate result of 3.95544. If a more concise answer is required, rounding becomes essential. Rounding to two decimal places, as demonstrated, adjusts the number to \(x \approx 3.96\). There are several ways to round numbers, but common strategies include:

• Rounding up if the digit following your rounding place is 5 or greater.
• Rounding down if the digit is less than 5.

Rounding ensures the solution is both simple to read and use while still accurately representing the calculated value to a necessary degree of precision.