Question

Difficulty: MediumLogarithms and Change of Base

If log2x+log4x+log16x=7\log_2 x + \log_4 x + \log_{16} x = 7, what is the value of xx?

  1. 1616Answer
  2. B
    44
  3. C
    22
  4. D
    88

Answer

1616
Using the change of base property, log4x=12log2x\log_4 x = \frac{1}{2}\log_2 x and log16x=14log2x\log_{16} x = \frac{1}{4}\log_2 x. Combining like terms yields (1+12+14)log2x=74log2x\left(1 + \frac{1}{2} + \frac{1}{4}\right)\log_2 x = \frac{7}{4}\log_2 x. Setting 74log2x=7\frac{7}{4}\log_2 x = 7 gives log2x=4\log_2 x = 4, which leads to x=24=16x = 2^4 = 16.

Step-by-Step Solution

1
Apply the change of base formula to express all logarithmic terms in base 2.
log4x=log2xlog24=12log2x\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{1}{2}\log_2 x and log16x=log2xlog216=14log2x\log_{16} x = \frac{\log_2 x}{\log_2 16} = \frac{1}{4}\log_2 x.
Logarithms with different bases must be converted to a common base to combine them.
2
Substitute the transformed terms back into the equation and factor out log2x\log_2 x.
\log_2 x + \frac{1}{2}\log_2 x + \frac{1}{4}\log_2 x = \left(1 + \frac{1}{2} + \frac{1}{4}\right)\log_2 x = \frac{7}{4}\log_2 x.
Factoring allows summing the coefficients of log2x\log_2 x.
3
Solve for log2x\log_2 x by equating the simplified expression to 7.
\frac{7}{4}\log_2 x = 7 \implies \log_2 x = 7 \times \frac{4}{7} = 4.
Isolating the logarithmic term gives its numerical value.
4
Convert the logarithmic equation to exponential form to solve for xx.
x = 2^4 = 16.
By definition, logba=c    a=bc\log_b a = c \iff a = b^c.

Key Concept

Change of Base Formula for Logarithms
Estimated Time:1m 30s
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