In question 60, the base change property was used.
This property says that:
[tex]log_b(c) = \frac{log_d(c)}{log_d(b)}[/tex]
Remember that the base of the neperian logarithm (ln) is the number of euler "e" [tex]ln(a) = log_e(a)[/tex]
So:
[tex]log_9(64) = \frac{ln(64)}{ln(9)}[/tex]
Option C
In question 62, use the base change property again. We have:
[tex]log_{1/2}(x ^ 2)[/tex]
Let's change the [tex]log_{1/2}[/tex] to the [tex]log_2.[/tex]
So:
[tex]log_{1/2}(x ^ 2) = \frac{log_2(x ^ 2)}{log_2 (1/2)}[/tex] ----------- lower the exponent in the numerator [tex](log_a(x ^ k) = klog_a( x)[/tex]) and calculate the log of the denominator.
[tex]= \frac{2log_2(|x|)}{- 1}[/tex]
[tex]log_{1/2}(x ^ 2) = -2log_2(|x|)[/tex]
Answer:
Option E
