Question

If \(\log _{3} 2 \log _{3}\left(2^{x}-5\right)\) and \(\log _{3}\left(2^{r}-7 / 2\right)\) are in AP then \(x\) is equal to : (a) 2 (b) 3 (c) 4 (d) 5

Short answer

In conclusion, based on the step-by-step solution, the value of \(x\) that makes the given expressions an Arithmetic Progression is \(x = 3\). The correct answer is (b) 3.

Step-by-step solution

Express the given terms as an AP

Since the three terms are in AP, the average of the first and third terms must be equal to the middle term. Write down the average expression: \[\frac{1}{2}\left(\log_{3}2 + \log_{3}\left(2^{r} - \frac{7}{2}\right)\right) = \log_{3}\left(2^{x} - 5\right)\]

Simplify and apply logarithmic properties

Apply the logarithmic property \(\log_{a}b + \log_{a}c = \log_{a}(bc)\) on the left side: \[\log_{3}\left(2\left(2^{r} - \frac{7}{2}\right)\right) = \log_{3}\left(2^{x} - 5\right)\] Now, set the arguments of the logarithms equal to each other since their bases are the same (3): \[2\left(2^{r} - \frac{7}{2}\right) = 2^{x} - 5\]

Solve for x

Expand the expression on the left side and simplify the equation: \[2^{r+1} - 7 = 2^{x} - 5\] Add 5 to both sides of the equation: \[2^{r+1} - 2 = 2^{x}\] Since we know that the terms are in an AP, we can assume that \(r = x - 1\) (an equal common difference between terms). Substitute this value in the equation: \[2^{x} - 2 = 2^{x}\] Due to the presence of \(2^{x}\) on both sides, we can conclude that \(2^{x} - 2^{x} = 0\). Since this equation is true, we move on to the next step.

Choose the correct value of x from the given options

Now, we need to check which value of \(x\) (from the given options) fits the general expression found above. Trying all the options: (a) \(x=2\): \(\ 2^{2} - 2 = 2 \neq 0\) (b) \(x=3\): \(\ 2^{3} - 2^{3} = 0\) (c) \(x=4\): \(\ 2^{4} - 2^{4} = 0\) (d) \(x=5\): \(\ 2^{5} - 2^{5} = 0\) All (b), (c), and (d) satisfy the expression, but only one value can be correct. Therefore, we need to refer back to the original AP expression. Substitute the values for \(x\): - For \(x=3\): \(\log_{3}\left(2^{3} - 5\right) = \log_{3}3\), which is true. - For \(x=4\): \(\log_{3}\left(2^{4} - 5\right) = \log_{3}11\), which is not in AP. - For \(x=5\): \(\log_{3}\left(2^{5} - 5\right) = \log_{3}27\), which is not in AP.

Conclusion

From our analysis, only the value \(x = 3\) makes the original equation an AP. Therefore, the answer is (b) 3.

Key concepts

Arithmetic Progression

An arithmetic progression (AP) is one of the simplest types of number sequences where a constant number, known as the common difference, is added to each term to get the next term. The general form of an AP is given by:

\[\begin{equation}a, a+d, a+2d, a+3d, \text{ ... }\text{ where }a\text{ is the first term and }d\text{ is the common difference.}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text\/a,\text{ }a+d,\text{ }a+2d,\text{ }a+3d,\text{ }...-\end{equation}\]
In an arithmetic series, it is possible to find any term given the first term and the common difference. This property helps solve many algebraic problems, such as the one posed in the exercise, where logarithmic terms are arranged to conform to an AP's rules.

Logarithm Properties

Logarithms are mathematical tools used to solve exponential equations and can be difficult to grasp at first. However, understanding their properties can make using them much more straightforward. Here are some of the fundamental properties of logarithms that are essential for our context:

• The product rule: \(\log_{a}(mn) = \log_{a}m + \log_{a}n\)
• The quotient rule: \(\log_{a}\frac{m}{n} = \log_{a}m - \log_{a}n\)
• Change of base formula: \(\log_{a}b = \frac{\log_{c}b}{\log_{c}a}\) for any positive base c
• The power rule: \(\log_{a}m^{n} = n\log_{a}m\)

Using these properties, we can combine, expand, or simplify logarithmic expressions to solve for unknown variables. The solution to the given problem demonstrates the use of the product rule to combine two logarithmic terms into a single term. Recognizing what logarithmic rule to apply at each step is a key to solving the more complicated logarithmic equations.

Solving Exponential Equations

Exponential equations feature variables in the exponent and can often be solved through the use of logarithms. The basic form of an exponential equation is \(a^{x} = b\), where 'a' is the base, 'x' is the exponent, and 'b' is the result. Solving these types of equations typically involves isolating the variable and sometimes applying logarithms to both sides. For instance, taking the logarithm of both sides with a base of 'a' yields \(\log_{a}(a^{x}) = \log_{a}(b)\), simplifying to \(x = \log_{a}(b)\) because of the inverse relationship between logarithms and exponents.

In our exercise, we have a slightly more complex scenario where we are required to recognize that the given logarithmic expressions form an arithmetic progression only if the exponent 'x' satisfies certain conditions. Here, the understanding of how to manipulate exponential and logarithmic expressions is crucial in deriving the appropriate value of 'x'. When solving exponential equations, patience and practice in recognizing when to apply logarithms and algebraic manipulation are key in finding the solution.