Problem 1 :
Solve for x :
log_{2}x = 1/2
Solution :
log_{2}x = 1/2
Convert to exponential form.
x = 2^{1/2}
x = √2
Problem 2 :
Solve for x :
log_{1/5}x = 3
Solution :
log_{1/5}x = 3
Convert to exponential form.
x = (1/5)^{3}
x = 1^{3}/5^{3}
x = 1/125
Problem 3 :
Solve for y :
log_{3}y = -2
Solution :
log_{3}y = -2
Convert to exponential form.
y = 3^{-2}
y = 1/3^{2}
y = 1/9
Problem 4 :
Solve for x :
log_{x}125√5 = 7
Solution :
log_{x}125√5 = 7
Convert to exponential form.
125√5 = x^{7}
5 ⋅ 5 ⋅ 5 ⋅ √5 = x^{7}
Each 5 can be expressed as (√5 ⋅ √5).
Then,
√5 ⋅ √5 ⋅ √5 ⋅ √5 ⋅ √5 ⋅ √5 ⋅ √5 = x^{7}
√5^{7 }= x^{7}
Because the exponents are equal, bases can be equated.
x = √5
Problem 5 :
Solve for x :
log_{x}0.001 = -3
Solution :
log_{x}0.001 = -3
Convert to exponential form.
0.001 = x^{-3}
1/1000 = 1/x^{3}
Take reciprocal on both sides.
1000 = x^{3}
10^{3} = x^{3}
Because the exponents are equal, bases can be equated.
10 = x
Problem 6 :
Solve for x :
x + 2log_{27}9 = 0
Solution :
x + 2log_{27}9 = 0
x = -2log_{27}9
x = log_{27}9^{-2}
Convert to exponential form.
27^{x} = 9^{-2}
(3^{3})^{x }= (3^{2})^{-2}
3^{3x} = 3^{-4}
Because the bases are equal, exponents can be equated.
3x = -4
x = -4/3
Problem 7 :
If 2logx = 4log3, then find the value of x.
Solution :
2logx = 4log3
Divide each side by 2.
logx = (4log3)/2
logx = 2log3
logx = log3^{2}
logx = log9
x = 9
Problem 8 :
If 3x is equal to log(0.3) to the base 9, then find the value of x.
Solution :
From the information given, we have
3x = log_{9}(0.3)
Solve for x.
3x = log_{9}(1/3)
3x = log_{9}1 - log_{9}3
3x = 0 - log_{9}3
3x = -log_{9}3
3x = -1/log_{3}9
3x = -1/log_{3}3^{2}
3x = -1/2log_{3}3
3x = -1/2(1)
3x = -1/2
x = -1/6
Problem 9 :
Solve the following equation :
log_{4}(x + 4) + log_{4}8 = 2
Solution :
log_{4}(x + 4) + log_{4}8 = 2
Combine the two terms on the left side.
log_{4}[8 ⋅ (x + 4)] = 2
log_{4}(8x + 32) = 2
8x + 32 = 4^{2}
8x + 32 = 16
Subtract by 32 from both sides
8x = -16
Divide both sides by 8.
x = -2
Problem 10 :
Solve the following equation :
log_{6}(x + 4) - log_{6}(x - 1) = 1
Solution :
log_{6}(x + 4) - log_{6}(x - 1) = 1
Combine the two terms on the left side
log_{6}[(x + 4)/(x - 1)] = 1
(x + 4)/(x - 1) = 6^{1}
(x + 4)/(x - 1) = 6
x + 4 = 6(x - 1)
x + 4 = 6x - 6
Subtract 6x from both sides.
x - 6x + 4 = -6
-5x + 4 = -6
Subtract 4 from both sides.
-5x = -6 - 4
-5x = -10
Divide both sides by -5.
x = 2
Problem 11 :
Solve the following equation :
log_{2}x + log_{4}x + log_{8}x = 11/6
Solution :
log_{2}x + log_{4}x + log_{8}x = 11/6
(1/log_{x}2) + (1/log_{x}4) + (1/log_{x}8) = 11/6
(1/log_{x}2) + (1/log_{x}2^{2}) + (1/log_{x}2^{3}) = 11/6
(1/log_{x}2) + (1/2log_{x}2) + (1/3log_{x}2) = 11/6
(1/log_{x}2) (1 + 1/2 + 1/3) = 11/6
(1/log_{x}2)(11/6) = 11/6
1/log_{x}2 = 1
1 = log_{x}2
x = 2
Problem 12 :
Given that
logx = m + n
logy = m – n
Find the value of log(10x/y^{2}) in terms of m and n.
Solution :
log(10x/y^{2}) = log10x - 1ogy^{2}
= log10 + logx - 2logy
= 1 + logx - 2logy
Substitute.
= 1 + (m + n) - 2(m - n)
= 1 + m + n - 2m + 2n
= 1 - m + 3n
Problem 13 :
Given that
logx + logy = log(x + y)
Solve for y in terms of x.
Solution :
logx + logy = log(x + y)
Use the Product Rule of Logarithm on the left side.
log(xy) = log(x + y)
xy = x + y
Subtract y from both sides.
xy - y = x
Factor.
y(x - 1) = x
Divide both sides by (x - 1).
y = x/(x - 1)
Problem 14 :
Given that
log_{10}2 = x
log_{10}3 = y
Find the value of log_{10}1.2 in terms of x and y.
Solution :
= log_{10}1.2
= log_{10}(12/10)
= log_{10}12 - log_{10}10
= log_{10}(4 ⋅ 3) - 1
= log_{10}4 + log_{10}3 - 1
= log_{10}2^{2} + log_{10}3 - 1
= 2log_{10}2 + log_{10}3 - 1
= 2x + y - 1
Problem 15 :
Solve for x :
100^{√}^{x} = log_{2}1024
Solution :
100^{√}^{x }= log_{2}1024
100^{√}^{x }= log_{2}2^{10}
100^{√}^{x }= 10log_{2}2
100^{√}^{x }= 10(1)
100^{√}^{x }= 10
(10^{2})^{√}^{x }= 10
10^{2}^{√}^{x }= 10
10^{2}^{√}^{x }= 10^{1}
2√x = 1
Divide both sides by 2.
√x = 1/2
Square both sides.
(√x)^{2} = (1/2)^{2}
x = 1/4
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