How do you evaluate #\frac { \log _ { 2} 16- \log _ { \pi } 1} { \log _ { 4\sqrt { 2} } 32- \log 0.1}#?

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Answer 1 · 2017-03-27T01:35:40

The expression has a value of #4/3#.

Use #log_a x = logx/loga# to rewrite.

#(log16/log2 - log1/log pi)/(log32/(log4sqrt(2)) - log0.1/log10)#

We now make a few modifications to the form to make the expression easier to work with.

#(log16/log2 - log1/logpi)/(log32/logsqrt(4^2 *2) - log(1/10)/log10)#

Now write the fractions in common powers.

#(log(2^4)/log(2^1) - log1/logpi)/(log32/log(32^(1/2)) -log(10)^-1/log10)#

Now use the laws #log1 = 0# and #loga^n = nloga#.

#((4log2)/log2 - 0/logpi)/(log32/(1/2log32) - (-1log10)/log10)#

#4/(2 + 1)#

#4/3#

Hopefully this helps!