Question

Difficulty: HardLogarithmic and Exponential Expressions and Equations

If aa, bb, and cc are positive real numbers greater than 11 such that logb(a)=32\log_b(a) = \frac{3}{2} and logc(b)=43\log_c(b) = \frac{4}{3}, what is the value of loga(abc)\log_a(abc)?

  1. A
    13\frac{1}{3}
  2. 136\frac{13}{6}Answer
  3. C
    239\frac{23}{9}
  4. D
    33
  5. E
    92\frac{9}{2}

Answer

136\frac{13}{6}
The correct answer is obtained by expanding loga(abc)\log_a(abc) into loga(a)+loga(b)+loga(c)\log_a(a) + \log_a(b) + \log_a(c). Since loga(a)=1\log_a(a) = 1 and loga(b)\log_a(b) is the reciprocal of logb(a)\log_b(a), we find loga(b)=23\log_a(b) = \frac{2}{3}. Using the chain rule of change of base, we calculate loga(c)=loga(b)logb(c)=2334=12\log_a(c) = \log_a(b) \cdot \log_b(c) = \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2}. Adding these three parts together yields 1+23+12=1361 + \frac{2}{3} + \frac{1}{2} = \frac{13}{6}.

Step-by-Step Solution

1
Apply the logarithmic product rule to expand the target expression.
loga(abc)=loga(a)+loga(b)+loga(c)\log_a(abc) = \log_a(a) + \log_a(b) + \log_a(c)
The logarithm of a product is equal to the sum of the logarithms of its individual factors.
2
Evaluate the first two terms of the expanded expression.
loga(a)=1\log_a(a) = 1, and since logb(a)=32\log_b(a) = \frac{3}{2}, we have loga(b)=1logb(a)=23\log_a(b) = \frac{1}{\log_b(a)} = \frac{2}{3}.
The logarithm of a base to itself is always 11, and the base reciprocal property states that logx(y)=1logy(x)\log_x(y) = \frac{1}{\log_y(x)}.
3
Evaluate the third term loga(c)\log_a(c) using base properties and the change of base formula.
Since logc(b)=43\log_c(b) = \frac{4}{3}, we have logb(c)=34\log_b(c) = \frac{3}{4}. Then, loga(c)=loga(b)logb(c)=2334=12\log_a(c) = \log_a(b) \cdot \log_b(c) = \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2}.
By applying the change of base formula, we can rewrite loga(c)\log_a(c) in terms of base bb as logb(c)logb(a)=loga(b)logb(c)\frac{\log_b(c)}{\log_b(a)} = \log_a(b) \cdot \log_b(c).
4
Sum the three evaluated logarithmic terms together.
1+23+12=66+46+36=1361 + \frac{2}{3} + \frac{1}{2} = \frac{6}{6} + \frac{4}{6} + \frac{3}{6} = \frac{13}{6}
Combining the values of the individual terms gives the final value of the expanded expression.

Key Concept

Properties of Logarithms and Change of Base Formula
Estimated Time:2m 0s
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