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Can someone explain how to do 60 and 62?

Can someone explain how to do 60 and 62?-example-1
User Carline
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1 Answer

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In question 60, the base change property was used.

This property says that:


log_b(c) = (log_d(c))/(log_d(b))

Remember that the base of the neperian logarithm (ln) is the number of euler "e"
ln(a) = log_e(a)

So:


log_9(64) = (ln(64))/(ln(9))

Option C


In question 62, use the base change property again. We have:


log_(1/2)(x ^ 2)

Let's change the
log_(1/2) to the
log_2.

So:


log_(1/2)(x ^ 2) = (log_2(x ^ 2))/(log_2 (1/2)) ----------- lower the exponent in the numerator
(log_a(x ^ k) = klog_a( x)) and calculate the log of the denominator.


= (2log_2(|x|))/(- 1)


log_(1/2)(x ^ 2) = -2log_2(|x|)

Answer:

Option E

User Dave Thieben
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