Module 4: Investiga Precalculus The Inverse of an Exponential The first property
ID: 3018573 • Letter: M
Question
Module 4: Investiga Precalculus The Inverse of an Exponential The first property of logarithms states log, (m n)- log, m+ log, . For example, log, 4+ loo.R r 2 is raised to in get a result of 8, and then add those two powers together (2+3 -5). We could have instcad log, (4-8)- log, (32) and determined directly that 5 is the power 2 must be raised to in order to result of 32 get a Use your understanding of logarithmic functions (knowledge of what the input and output represent) to justify at least one of the logarithmic properties above . Use the properties of logarithms to simplify the following expressions. a. In(x)+InCx) c. log.(16)-log.(4) b. log,(5) log, (2) e. log,9)+ log, (4)-log,(6) f. log, (49)-log.(-3) d. log, (2)+ log,(3) . Solve each of the following for x a. log, (30)-3 b. log(1.5x)=0 c. log, x+ log, (2x)-3 @iet /(x)-10, and g(x)=log(x). a. Complete the tables 0.01 0.1 10 100 Evaluate the following expressions: i. g(f(-2)) iv.f(g(10) What do you notice about the relationship between the functions fand g? b. i. gif(0) v.f(g(x)) vi. g(f(x)) c. 12. Use the fact that the word "log" is the name of a function and the statement log, (x) represeExplanation / Answer
9) a )
ln x + ln x
= ln (x*x) = ln (x^2 )
b) log 3 (5) + log 3 (2)
= log 3 (5*2) = log 3 (10 )
c) log 4 (16) - log 4 (4)
= log 4 ( 16/4 )
= log 4 ( 4)
= 1
d) log 3 (2) + log 5 (3 )
applying change of base property
(log 2 / log 3 )+ (log 3 / log 5 )
e) log 3 (9) + log 3 (4) - log 3 (6)
= log 3 (9*4 ) - log 3 (6)
log 3 ( 36 / 6 )
= log 3 ( 6 )
f) log 7 ( 49 ) - log 7 ( -3)
= log 7 ( 49 / -3 )
10) a) log 5 (30x^2) = 3
changing to exponential form
30x^2 = 5^3
30x^2 = 125
x^2 = 125/30
x = +- 2.0412
b) log (1.5x )= 0
10^0 = 1.5x
1 = 1.5x
x = 1/ 1.5 = 0.667
c) log 3 (x) + log 3 (2x ) = 3
log 3 ( 2x^2 ) = 3
2x^2 = 3^3
2x^2 = 27
x^2 = 27/2
x = sqrt ( 27/2 )
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