Evaluate \(log_{\sqrt{2}}4+log_{\frac{1}{2}}16-log_{4}32\)

  • A -5.5
  • B -2.5
  • C 2.5
  • D 5.5

The correct answer is B. -2.5

We can use the properties of logarithms to evaluate the expression \(log_{\sqrt{2}}4+log_{\frac{1}{2}}16-log_{4}32\). Recall that \(log_{a}b = \frac{log_{c}b}{log_{c}a}\) for any positive base c not equal to 1. Using this property, we can rewrite the expression as follows:

\(log_{\sqrt{2}}4+log_{\frac{1}{2}}16-log_{4}32\)

\(= \frac{log_{2}4}{log_{2}\sqrt{2}} + \frac{log_{2}16}{log_{2}\frac{1}{2}} - \frac{log_{2}32}{log_{2}4}\)

Since \(log_{a}a^b = b\) for any positive base a not equal to 1, we can simplify the expression further:

\(= \frac{2}{\frac{1}{2}} + \frac{4}{-1} - \frac{5}{2}\)

\(= 4 - 4 - \frac{5}{2}\)

\(= -\frac{5}{2}\)

So, the value of the expression \(log_{\sqrt{2}}4+log_{\frac{1}{2}}16-log_{4}32\) is -\(\frac{5}{2}\)

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