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For how many distinct real values of $$x$$ does the equation below hold true? (Consider $$a$$ > 0.)
$$\dfrac{x^2 \log_a(16)}{\log_a(32)} - \dfrac{\log_a(64)}{\log_a(32)} - x = 0 $$
In the given equation $$\dfrac{x^2 \log_a(16)}{\log_a(32)} - \dfrac{\log_a(64)}{\log_a(32)} - x = 0 $$
We know the property of log that $$\dfrac{\log_ab}{\log_ac}=\log_cb$$
So, $$\frac{\log_a16}{\log_a32}=\log_{32}16=\log_{2^5}2^4=\frac{4}{5}$$
Similarly, $$\frac{\log_a64}{\log_a32}=\log_{32}64=\log_{2^5}2^6=\frac{6}{5}$$
Hence, the equation can be written as
$$\frac{4}{5}x^2-\frac{6}{5}-x=0$$
or, $$4x^2-6-5x=0$$
Evaluating the discriminant we get
$$D=b^2-4ac=\left(-5\right)^2-4\left(4\right)\left(-6\right)$$
or, $$D=121>0$$
Hence, the equation has two distinct real roots.
But as the question mentions $$a>0$$, so it can take the value $$a=1$$, for which the log term will not be defined, hence there will be no defined solution.
Hence, the answer is Depends on the value of $$a$$
Yes, Logarithms, Surds, and Indices are important algebra topics in XAT Quantitative Ability. These concepts are frequently used in simplifying expressions and solving higher-level quantitative aptitude questions.
XAT may include questions on logarithmic properties, logarithmic equations, surd simplification, exponents, powers, and indices-based algebraic expressions.
Start by understanding the fundamental rules and properties of logarithms, surds, and indices. Regular practice of topic-wise questions and previous year papers can help improve conceptual clarity and speed.
The difficulty level varies from year to year. While some questions are formula-based, others require multiple concepts to be applied together. A strong grasp of basics makes these questions much easier to solve.
Cracku's XAT Logarithms, Surds & Indices Questions are designed according to the latest XAT exam pattern and difficulty level. They provide topic-wise questions, detailed solutions and shortcut methods to help aspirants strengthen concepts, improve accuracy, and solve these questions more efficiently.
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