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How do you expand #log_3 48#?

Answer 1

Lily Anderson

#log_3 48 = 1+4 log_3 2#

Note that #48 = 3*2^4#

So: #log_3 48 = log_3 (3*2^4) = log_3 3 + log_3 (2^4) = 1+4 log_3 2#

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Answer 2

Sophie Adams

To expand log base 3 of 48, we can use the property of logarithms that states log base a of (mn) is equal to log base a of m plus log base a of n:

log base 3 of 48 = log base 3 of (16 * 3) = log base 3 of 16 + log base 3 of 3

Now, since 16 is 4 raised to the power of 2, we can write:

log base 3 of 16 = log base 3 of (4^2) = 2 * log base 3 of 4

Putting it all together:

log base 3 of 48 = 2 * log base 3 of 4 + log base 3 of 3

Now, log base 3 of 4 can be further expanded using the property that log base a of b to the power of c is equal to c times log base a of b:

log base 3 of 4 = log base 3 of (2^2) = 2 * log base 3 of 2

Substitute this back into the equation:

log base 3 of 48 = 2 * (2 * log base 3 of 2) + log base 3 of 3

Finally, simplify:

log base 3 of 48 = 4 * log base 3 of 2 + log base 3 of 3

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.