Question

If \(\log _{144} 729=x\), then the value of \(\log _{36} 256\) is (1) \(\frac{4(3-x)}{(3+x)}\) (2) \(\frac{4(3+\mathrm{x})}{(3-\mathrm{x})}\) (3) \(\frac{(3+\mathrm{x})}{4(3-\mathrm{x})}\) (4) \(\frac{(3-\mathrm{x})}{4(3+\mathrm{x})}\)

Short answer

Short Answer: None of the given options match the value \(2\), therefore there is no correct answer for this exercise.

Step-by-step solution

Convert to the exponential form

We can rewrite the given expression in exponential form \(144^x = 729\)

Simplify the equation

Rewrite bases 144 and 729 as powers of 2 and 3: \((2^4 \cdot 3^2)^x = 3^6\) Now expand and simplify: \(2^{4x} \cdot 3^{2x} = 3^6\)

Write down 2 separate equations

Since the bases are equal (2 and 3), we can write down 2 separate equations with the same exponents: \(2^{4x} = 2^0\) and \(3^{2x} = 3^6\)

Solve for x

From the first equation, we can determine that: \(4x = 0\) so \(x = 0\) Now we know that the value of x is 0, we can use it to find the value of \(\log_{36} 256\).

Write down the expression for \(\log_{36} 256\)

We know that \(36=2^2\cdot 3^2\), and \(256=2^8\). Using this, we can write down the expression as: \(\log_{2^2\cdot3^2}2^8\)

Change of Base Formula

Using the change of base formula we get: \(\dfrac{\log_{2}2^8}{\log_{2} 2^2\cdot3^2}\) and simplify the expression: \(\dfrac{8}{2+2}=\dfrac{8}{4}=2\)

Convert \(2\) to the form of the given options

Now we want to express our result 2 using x in the given format. Since x=0, the correct format will be when the fraction equals 2. We can test given options: (1) \(\frac{4(3-0)}{(3+0)}=\frac{12}{3}=4\) (2) \(\frac{4(3+0)}{(3-0)}=\frac{12}{3}=4\) (3) \(\frac{(3+0)}{4(3-0)}=\frac{3}{12}=\frac{1}{4}\) (4) \(\frac{(3-0)}{4(3+0)}=\frac{3}{12}=\frac{1}{4}\) None of the given options match the value \(2\), therefore there is no correct answer for this exercise.

Key concepts

Exponential Form

Turning logarithmic equations into exponential form can make them much easier to visualize and solve. In our example, we start with the logarithmic equation \( \log_{144} 729 = x \). Converting this into exponential form, we have \( 144^x = 729 \). This form tells us that 144 raised to the power of \( x \) gives us 729. By expressing the equation in terms of powers of smaller numbers, like 2 and 3, we can make calculations more straightforward.

Change of Base Formula

The change of base formula is an essential tool in solving logarithmic problems when the bases of the logarithms are not convenient or the same. The formula allows us to express a logarithm in terms of logs of any other base. For instance, consider \( \log_{36} 256 \). Using the change of base formula, it becomes \( \frac{\log_{2} 256}{\log_{2} 36} \). This conversion helps in simplifying calculations, as you can switch to base 10 or base 2, which are generally more manageable using calculators.

Simplification

Simplifying expressions often requires breaking down components into their fundamental parts. For the problem at hand, we rewrite 144 and 729 as powers of smaller base numbers, \( 144 = (2^4 \cdot 3^2) \) and \( 729 = 3^6 \). This helps in equating powers and managing separate equations. Simplification further aids the process as it reduces the equation into simpler pieces, making it easier to handle and solve. This step is crucial to avoid unnecessary complexity.

Powers and Exponents

Understanding powers and exponents is a fundamental aspect of working with logarithms and exponential equations. In the problem, we determine \( 144 = (2^4 \cdot 3^2) \). By expanding, we have \( 144^x = (2^{4x} \cdot 3^{2x}) \), which equals \( 3^6 \). It's important to recognize how exponents distribute over multiplication and how to equate them when the bases match. Mastery of powers and exponents allows for efficient manipulation of equations, letting us solve for unknowns like \( x \) seamlessly.