Problem 10 - Algebra 2

Question

Solve each equation. Round to the nearest ten-thousandth. Check your answers. $$ 12^{y-2}=20 $$

Step-by-Step Solution

Answer

The solution to the equation, rounded to the nearest ten-thousandth, is y ≈ 2.8793

Step 1: Convert into Logarithmic Form

To convert the equation into a logarithmic form, take the natural log (ln) on both sides of the equation. The equation then becomes: \(ln(12^{y-2})=ln(20)\). Then apply the power rule of logarithms which says that \(\ln(a^b)=b*ln(a)\). So the equation turns into: \((y-2)*ln(12)=ln(20)\).

Step 2: Solve for y

To solve for 'y', first expand the left side of the equation to isolate 'y'. So the equation becomes: \(y*ln(12)-2*ln(12)=ln(20)\). Then add \(2*ln(12)\) to both sides of the equation to isolate 'y' on the left side. So the equation becomes: \(y*ln(12)=ln(20)+2*ln(12)\). Finally, divide by \(ln(12)\) from both sides. The equation becomes: \(y=(ln(20)+2*ln(12))/ln(12)\). Calculate the right side of the equation with the help of a calculator to get the approximate value of 'y'. So the value of 'y' is approximately 2.8793 (rounded to the nearest ten-thousandth).

Step 3: Check your Answer

To verify that this is the correct answer, substitute 'y' with 2.8793 (the found value) into the original equation. Perform the equation computations and check if the result equals 20.

Key Concepts

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. They are incredibly useful when dealing with equations where the variable is in the exponent, as they help transform them into a more manageable linear form. Here, you are required to solve an equation of the form \(12^{y-2}=20\). To tackle this, you use logarithmic functions to peel away the exponent.The process begins by applying a logarithm to both sides of the equation. You can choose any logarithm, but the most common are base 10 (log) and the natural logarithm (ln), which is base \(e\). This will transform the equation:

• From: \(12^{y-2} = 20\)
• To: \(\ln(12^{y-2}) = \ln(20)\)

Now, use the power rule of logarithms, which states that \(\ln(a^b) = b*\ln(a)\), to simplify the left-hand side into a linear form:

• \((y - 2) * \ln(12) = \ln(20)\)

This transformation allows us to solve for the variable \(y\) as a linear problem, showcasing the power of logarithmic functions for equations involving exponential expressions.

Natural Logarithm

The natural logarithm, often abbreviated as ln, is a logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. It is widely used in calculus and mathematical modeling, and it is particularly effective for solving exponential equations.When solving exponential equations like \(12^{y-2}=20\), employing the natural logarithm can simplify the process vastly. The steps to using the natural logarithm are straightforward:

• First, take the natural logarithm of both sides to manipulate the exponent.
• Apply the power rule, which helps move the exponent down as a coefficient in front of the logarithm: \(\ln(12^{y-2}) = \ln(20)\) becomes \((y-2) * \ln(12) = \ln(20)\).

By re-arranging the equation, you isolate \(y\), making it possible to solve algebraically. This method not only finds the value of \(y\) accurately but also highlights the natural logarithm's efficiency in handling problems involving exponential growth or decay.Once solved, check the result by substituting back to ensure that the original equation holds true for your computed value of \(y\). This ensures a complete understanding and mastery over both the solution and the underlying mathematical concepts.

Rounding Numbers

Rounding is a mathematical technique to simplify numbers, making them easier to work with, usually while maintaining an approximation close to the original value. In your exercise, you are asked to round the answer to the nearest ten-thousandth, which means keeping four decimal places.Once you've solved the equation for \(y\), for example, \(y = 2.879278\), you'll round this to \(2.8793\), the nearest ten-thousandth. The rules for rounding are consistent:

• Identify the required place value, the fourth decimal in this case.
• Look at the digit immediately to the right; if it’s 5 or greater, increase the last digit by one.
• If it's less than 5, keep the last digit as is.

Rounding helps maintain precision without overwhelming you with unnecessary digits. It's a critical skill, particularly in science and engineering, where significant figures play a vital role.After rounding your answer, verify if replacing \(y\) with the rounded value satisfies the original equation. This step assures the solution's accuracy and reliability, teaching the importance of both precise calculation and thoughtful approximation.