Home > Precalculus > Logarithm-- Inverse of an Exponential Function

How do you evaluate #10 log18 - log3#?

Answer 1

Sadie Ackerman

#10log(18)-log(3)=log(18^10/3)#

Remember

Therefore #color(white)("XXX")10log(18)-log(3)#

#color(white)("XXX")=log(18^10)-log(3)#

#color(white)("XXX")=log(18^10/3)#

This could be evaluated using a calculator as: #color(white)("XXX")=12.0756#

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Answer 2

James Allen

To evaluate (10 \log_{18} - \log_3), we can use the properties of logarithms.

[10 \log_{18} - \log_3]

Using the property (n \log_a(b) = \log_a(b^n)), we can rewrite the expression:

[= \log_{18}(18^{10}) - \log_3(1)]

Since (18^{10}) equals (3^{20}) (since (18 = 3^2) and (10 \times 2 = 20)), the expression simplifies to:

[= \log_{18}(3^{20}) - \log_3(1)]

Now, using the property (\log_a(b^c) = c \cdot \log_a(b)), we can rewrite the expression:

[= 20 \log_{18}(3) - 0]

[= 20 \log_{18}(3)]

This cannot be simplified further without a calculator that can handle logarithms with bases other than 10 or e. So, (20 \log_{18}(3)) is the final answer.

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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.