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How do you solve #\log _ { 8} \frac { 23} { 18}#?

Answer 1

#log_8(23)-2log_8(3)-1/3#

#log_8(23/18)#

Expand as much as possible: #=log_8(23)-log_8(18)# #=log_8(23)-log_8(3*3*2)# #=log_8(23)-[log_8(3^2)+log_8(2)]# #=log_8(23)-[2log_8(3)+1/3]#

#=log_8(23)-2log_8(3)-1/3#

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Answer 2

Ariana Alvarez

To solve ( \log_8 \frac{23}{18} ), you can use the change of base formula for logarithms, which states that ( \log_b x = \frac{\log_c x}{\log_c b} ). Let's use base 10 as the new base (c), so the expression becomes:

[ \log_8 \frac{23}{18} = \frac{\log_{10} \frac{23}{18}}{\log_{10} 8} ]

Now, calculate the logarithms using a calculator:

[ \log_{10} \frac{23}{18} \approx 0.0913 ] [ \log_{10} 8 = \frac{\log_{10} 2^3}{\log_{10} 10} = \frac{3 \log_{10} 2}{1} ]

Next, substitute these values into the equation:

[ \frac{0.0913}{3 \log_{10} 2} ]

Finally, calculate ( \log_{10} 2 ) and divide to get the result:

[ \log_{10} 2 \approx 0.3010 ] [ \frac{0.0913}{3 \times 0.3010} \approx \frac{0.0913}{0.9030} \approx 0.1012 ]

So, ( \log_8 \frac{23}{18} \approx 0.1012 ).

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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.