Question

Find the value of \(\frac{\log \sqrt{27}+\log \sqrt{8}-\log \sqrt{125}}{\log 6-\log 5}\) : (a) \(\frac{2}{3}\) (b) \(\frac{1}{3}\) (c) \(\frac{3}{2}\) (d) none of these

Short answer

\(\frac{\log \sqrt{27}+\log \sqrt{8}-\log \sqrt{125}}{\log 6-\log 5}\) a) \(2\) b) \(\frac{5}{2}\) c) \(\frac{3}{2}\) d) \(\frac{3}{5}\) Answer: (c) \(\frac{3}{2}\)

Step-by-step solution

Rewrite expression using power rule

Rewrite \(\log \sqrt{27}\), \(\log \sqrt{8}\), \(\log \sqrt{125}\) and rearrange the terms using the quotient rule. The expression becomes: \[\frac{\frac{1}{2}\log 27+\frac{1}{2}\log 8-\frac{1}{2}\log 125}{\log 6-\log 5}\]

Apply quotient rule

Apply the quotient rule \(\log a - \log b = \log \frac{a}{b}\) to the numerator and denominator. The expression becomes: \[\frac{\log \frac{27 \cdot 8}{125}}{\log \frac{6}{5}}\]

Simplify the numbers in the expression

Simplify the numbers inside the logarithms. The expression becomes: \[\frac{\log \frac{216}{125}}{\log \frac{6}{5}}\]

Apply change of base formula

Apply the change of base formula \(\log_a {b}=\frac{\log_c {b}}{\log_c{a}}\) to the logarithmic expression with a new base \(c\). The expression becomes: \[\frac{\log_c \frac{216}{125}}{\log_c \frac{6}{5}}\] Observing that any base can be used to evaluate the expression, we can use a base convenient to simplify the fraction.

Cancel common factors

Cancel the common factors in the fraction. The expression becomes: \[\frac{\log \frac{216}{125}}{\log \frac{6}{5}} = \frac{3}{2}\] By simplifying the expression and comparing with the given options, we find that the value is:

Answer

\(\frac{3}{2}\) which corresponds to option (c).

Key concepts

Logarithmic Expressions

Logarithmic expressions involve the logarithm function, which is the inverse operation of exponentiation. In simpler terms, if you have an equation like \( b^x = y \), then the corresponding logarithmic form would be \( \text{log}_b(y) = x \). Here, \( b \) is called the base of the logarithm, \( y \) is the argument or input, and \( x \) represents the logarithm of \( y \) to the base \( b \).
When dealing with logarithmic expressions, it's essential to understand the properties of logarithms, such as the product rule (\( \text{log}_b(mn) = \text{log}_b(m) + \text{log}_b(n) \)), the quotient rule, and the power rule (\( \text{log}_b(m^k) = k \text{log}_b(m) \)). These properties help us simplify complex logarithmic expressions and solve logarithmic equations.
For instance, in the given exercise, square roots can be interpreted as numbers raised to the \( \frac{1}{2} \) power, allowing us to use the power rule to simplify the terms inside the logarithms. Transforming an expression into its logarithmic components can set a clear path toward problem-solving, especially when working with multiplication, division, or powers inside a logarithm.

Quotient Rule of Logarithms

The quotient rule of logarithms is a fundamental property that states that the logarithm of a quotient is equal to the difference of the logarithms of the numerator and the denominator: \( \text{log}_b\frac{m}{n} = \text{log}_b(m) - \text{log}_b(n) \).
This rule is applicable for any base \( b \) that's a positive real number, excluding 1, and for any positive real numbers \( m \) and \( n \). Simplifying logarithmic expressions using the quotient rule is critical for breaking down more complex terms into manageable pieces. This approach is demonstrated in our exercise where the logarithms in the numerator and the denominator were combined using the quotient rule, transforming the complex initial expression into a more straightforward comparative log-ratio.
Understanding the quotient rule not only assists in simplifications but also aids in understanding the relationship between exponents and logarithms, since it mirrors the exponent rule of dividing powers with the same base. Such conceptual parallels can often help students better grasp the underlying principles of logarithms.

Simplification of Logarithmic Expressions

Simplification of logarithmic expressions, as seen in the step-by-step problem-solving process, is about reducing the complexity and making expressions manageable. This process involves applying logarithmic rules such as the product rule, the quotient rule, and the power rule to combine or break apart logarithms.
Additionally, simplification might include basic arithmetic within the arguments of the logarithms and sometimes changing the base, as was demonstrated in the use of the change of base formula (Step 4 & 5). The change of base formula expresses a logarithm in terms of logarithms with a different base: \( \text{log}_a b = \frac{\text{log}_c b}{\text{log}_c a} \) for bases \( a \) and \( c \) that are positive and not equal to 1. This strategy can be particularly beneficial when the logs in the expression involve awkward numbers or when no common base exists.
In the given exercise, after applying the quotient rule, simplifying the numbers inside the logarithms to \( \frac{216}{125} \) and \( \frac{6}{5} \), and then canceling out the logs using the change of base formula, we were able to reach the simplified solution \( \frac{3}{2} \). It helped students make the connections between operations and arrive at the correct answer, advancing their understanding of how to manipulate and simplify logarithmic expressions effectively.