Question

Difficulty: Very hardLogarithms and Change of Base

Given that logab=2\log_a b = 2 and logbc=3\log_b c = 3, what is the value of logabc(a3b2c)\log_{a b c} (a^3 b^2 c)?

  1. 139\frac{13}{9}Answer
  2. B
    32\frac{3}{2}
  3. C
    119\frac{11}{9}
  4. D
    116\frac{11}{6}

Answer

139\frac{13}{9}
Using the change of base chain rule, logac=logablogbc=2×3=6\log_a c = \log_a b \cdot \log_b c = 2 \times 3 = 6. Changing the base of the target expression to base aa gives loga(a3b2c)loga(abc)\frac{\log_a (a^3 b^2 c)}{\log_a (a b c)}. Expanding both terms using product and power rules gives numerator 3(1)+2(2)+6=133(1) + 2(2) + 6 = 13 and denominator 1+2+6=91 + 2 + 6 = 9, resulting in 139\frac{13}{9}.

Step-by-Step Solution

1
Express logac\log_a c using the change of base relationship.
logac=logablogbc=2×3=6\log_a c = \log_a b \cdot \log_b c = 2 \times 3 = 6
By the change of base rule (chain rule of logarithms), logablogbc=logac\log_a b \cdot \log_b c = \log_a c.
2
Apply the change of base formula to convert logabc(a3b2c)\log_{a b c} (a^3 b^2 c) to base aa.
logabc(a3b2c)=loga(a3b2c)loga(abc)\log_{a b c} (a^3 b^2 c) = \frac{\log_a (a^3 b^2 c)}{\log_a (a b c)}
The change of base formula states that logBX=logaXlogaB\log_B X = \frac{\log_a X}{\log_a B}.
3
Expand the numerator and denominator using logarithmic product and power rules.
Numerator: 3logaa+2logab+logac=3(1)+2(2)+6=133\log_a a + 2\log_a b + \log_a c = 3(1) + 2(2) + 6 = 13. Denominator: logaa+logab+logac=1+2+6=9\log_a a + \log_a b + \log_a c = 1 + 2 + 6 = 9.
loga(XYZ)=logaX+logaY+logaZ\log_a (X Y Z) = \log_a X + \log_a Y + \log_a Z and loga(Xk)=klogaX\log_a (X^k) = k \log_a X.
4
Divide the expanded numerator by the denominator.
139\frac{13}{9}
Substituting the computed values yields the simplified fraction 139\frac{13}{9}.

Key Concept

Change of Base Rule and Logarithmic Expansion Laws
Estimated Time:2m 0s
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