- b.Finding xsuch that log 5 1 = xis equivalent
to finding xsuch that 5x= 1. We conclude that
the solution of this equation is 0. - d.Finding xsuch that log 16 64 = xis equivalent
to finding xsuch that 16x= 64. We rewrite the
expressions on both sides of the equation
using the same base of 2. Indeed, observe that
16 x= (2^4 )x= 2^4 xand 64 = 2^6. Hence, the value
ofxwe seek is the solution of the equation 4x
= 6, which is x= ^64 = ^32 . - c.The equation log 6 x= 2 is equivalent to
x= 6^2 = 36. So, the solution is x= 36. - b.logax= loga(5a) = loga5+loga(a) =
loga5 + loga(a) = loga5+^12 logaa- b.log 3 (3^4 93 ) = log 3 (3^4 ) + log 3 (9^3 )=
4log 3 3 + 3log 3 9 = 4(1) +3(2) = 10- d.The equation 5^3 x–1= 7 is equivalent to
log 5 7 = 3 – 1. This equation is solved as
follows:
log 5 7 = 3x– 1
3 x= 1 + log 5 7
x= ^13 (1 + log 5 7)- b.The given expression can be written as one
involving the terms logaxand logay:
logayx 3 = logax– loga(y^3 ) = logax– 3logay
Substituting logax= 2 and logay= –3 into this
expression yieldslogayx 3 = logax– 3loga(y) = 2 – 3(–3) =
2 + 9 = 11- b.The equation 3log 32 = x is equivalent to log 3 x
= log 3 2. So, the solution is x= 2. - c.Since f(x) = logaxand g(x) = axare inverses,
it follows by definition that f(g(x)) = loga(ax)
= x.
798. b.First, write the expression on the left side as
the ln of a single expression.
3ln^1 x= ln8
ln^1 x
3
= ln81
x3
= 8x–3= 8
(x–3)= 8
x= ^12 - b. e ln3= eln(3 )= 3 = =
- c.The given expression can be written as one
involving the terms ln xand ln y:
ln =
ln (e^2 y) – ln(x) =
ln (e^2 ) + ln y– ln(x ) =
2ln(e) + ln y– ^12 ln x=
2 + ln y– ^12 ln xSubstituting ln x= 3 and ln y= 2 into this
expression yieldsln ( ) = 2 + ln y– ^12 ln x= 2 + 2 – ^12 (3) =4 – ^32 = ^52 e^2 y
x^12e^2 y
x^3
3
^1
3 –1 2 –1 2 –1 2–1 3 –1 3^12ANSWERS & EXPLANATIONS–