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1 answer:
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The correct question is
Using the graph of f(x) = log2x below, approximate the value of y in the equation 2^(2y) = 5
we have
2^(2y) = 5--------------> applying base 2 logarithm both members
2y*log2(2)=log2(5)
log2(2)=1
then
2y=log2(5)
y=[log2(5)]/2
using the graph
for x=5 the approximate value of log2(5) is 2.3
see the attached figure
so
y=[log2(5)]/2--------> y=[2.3]/2----------> y=1.15
the answer is
the approximate value of y is 1.15
Using the graph of f(x) = log2x below, approximate the value of y in the equation 2^(2y) = 5
we have
2^(2y) = 5--------------> applying base 2 logarithm both members
2y*log2(2)=log2(5)
log2(2)=1
then
2y=log2(5)
y=[log2(5)]/2
using the graph
for x=5 the approximate value of log2(5) is 2.3
see the attached figure
so
y=[log2(5)]/2--------> y=[2.3]/2----------> y=1.15
the answer is
the approximate value of y is 1.15