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Let’s check out your problem:
6
6
6
\newline
4
4
4
(a) Given
log
2
3
=
p
\log _{2} 3=p
lo
g
2
3
=
p
and
log
2
5
=
q
\log _{2} 5=q
lo
g
2
5
=
q
, express and simplify
log
8
2
5
+
log
3
2
\log _{8} \frac{2}{5}+\log _{3} 2
lo
g
8
5
2
+
lo
g
3
2
in terms of
p
p
p
and
q
q
q
.
\newline
[
3
3
3
]
View step-by-step help
Home
Math Problems
Precalculus
Quotient property of logarithms
Full solution
Q.
6
6
6
\newline
4
4
4
(a) Given
log
2
3
=
p
\log _{2} 3=p
lo
g
2
3
=
p
and
log
2
5
=
q
\log _{2} 5=q
lo
g
2
5
=
q
, express and simplify
log
8
2
5
+
log
3
2
\log _{8} \frac{2}{5}+\log _{3} 2
lo
g
8
5
2
+
lo
g
3
2
in terms of
p
p
p
and
q
q
q
.
\newline
[
3
3
3
]
Change Base to Base
2
2
2
:
Change the base of
log
8
\log_{8}
lo
g
8
to base
2
2
2
using the change of base formula:
log
b
a
=
log
k
a
log
k
b
\log_{b} a = \frac{\log_{k} a}{\log_{k} b}
lo
g
b
a
=
l
o
g
k
b
l
o
g
k
a
.
log
8
(
2
5
)
=
log
2
(
2
5
)
log
2
8
.
\log_{8} \left(\frac{2}{5}\right) = \frac{\log_{2} \left(\frac{2}{5}\right)}{\log_{2} 8}.
lo
g
8
(
5
2
)
=
l
o
g
2
8
l
o
g
2
(
5
2
)
.
Simplify Log Base
2
2
2
of
8
8
8
:
Simplify
log
2
8
\log_{2} 8
lo
g
2
8
since
8
8
8
is
2
2
2
cubed,
log
2
8
\log_{2} 8
lo
g
2
8
equals
3
3
3
.
\newline
log
8
(
2
5
)
=
log
2
(
2
5
)
3
.
\log_{8} \left(\frac{2}{5}\right) = \frac{\log_{2} \left(\frac{2}{5}\right)}{3}.
lo
g
8
(
5
2
)
=
3
l
o
g
2
(
5
2
)
.
Split Log Base
2
2
2
of
(
2
5
)
(\frac{2}{5})
(
5
2
)
:
Split log base
2
2
2
of
(
2
5
)
(\frac{2}{5})
(
5
2
)
into log base
2
2
2
of
2
2
2
minus log base
2
2
2
of
5
5
5
using the quotient rule.
\newline
log
8
(
2
5
)
=
(
log
2
2
−
log
2
5
)
/
3
\log_{8}(\frac{2}{5}) = (\log_{2} 2 - \log_{2} 5) / 3
lo
g
8
(
5
2
)
=
(
lo
g
2
2
−
lo
g
2
5
)
/3
.
Substitute Given Values:
Substitute the given values:
log
2
3
=
p
\log_{2} 3 = p
lo
g
2
3
=
p
and
log
2
5
=
q
\log_{2} 5 = q
lo
g
2
5
=
q
.
\newline
log
8
(
2
5
)
=
1
−
q
3
\log_{8} \left(\frac{2}{5}\right) = \frac{1 - q}{3}
lo
g
8
(
5
2
)
=
3
1
−
q
because
log
2
2
=
1
\log_{2} 2 = 1
lo
g
2
2
=
1
.
Change Base to Base
2
2
2
:
Change the base of
log
3
2
\log_{3} 2
lo
g
3
2
to base
2
2
2
using the change of base formula.
log
3
2
=
log
2
2
log
2
3
\log_{3} 2 = \frac{\log_{2} 2}{\log_{2} 3}
lo
g
3
2
=
l
o
g
2
3
l
o
g
2
2
.
Substitute Given Value:
Substitute the given value for
log
2
3
\log_{2} 3
lo
g
2
3
.
\newline
log
3
2
=
1
p
\log_{3} 2 = \frac{1}{p}
lo
g
3
2
=
p
1
because
log
2
2
=
1
\log_{2} 2 = 1
lo
g
2
2
=
1
.
Combine Expressions:
Combine the two expressions.
\newline
log
8
(
2
5
)
+
log
3
2
=
1
−
q
3
+
1
p
\log_{8}\left(\frac{2}{5}\right) + \log_{3}2 = \frac{1 - q}{3} + \frac{1}{p}
lo
g
8
(
5
2
)
+
lo
g
3
2
=
3
1
−
q
+
p
1
.
Simplify Expression:
Simplify the expression.
\newline
Final expression in terms of
p
p
p
and
q
q
q
:
1
−
q
3
+
1
p
\frac{1 - q}{3} + \frac{1}{p}
3
1
−
q
+
p
1
.
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\newline
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\newline
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\newline
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=
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\newline
Choices:
\newline
(A)
g
(
t
)
g(t)
g
(
t
)
decreases by
3
3
3
\newline
(B)
g
(
t
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(
t
)
increases by
3
3
3
\newline
(C)
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(
t
)
g(t)
g
(
t
)
decreases by
3
%
3\%
3%
\newline
(D)
g
(
t
)
g(t)
g
(
t
)
increases by
200
%
200\%
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f(x)
f
(
x
)
increases by
6
6
6
over every unit interval in
x
x
x
and
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0
)
=
0
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)
=
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\newline
Which could be a function rule for
f
(
x
)
f(x)
f
(
x
)
?
\newline
Choices:
\newline
(A)
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x
)
=
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x
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x
)
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x
\newline
(B)
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x
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x
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x
)
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x
)
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x
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x
)
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1
\newline
(D)
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(
x
)
=
x
6
f(x) = \frac{x}{6}
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(
x
)
=
6
x
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