Problems 1-5 : Solve for x.
Problem 1 :
3^{x} = 27
Problem 2 :
2^{x - 5} = ¹⁄₁₆
Problem 3 :
6 ⋅ 3^{x} = 54
Problem 4 :
4^{x }= 7(2^{x}) + 18
Problem 5 :
5^{x }= (√25)^{-7 }⋅ (√5)^{-5})
Problems 6-10 : Solve for x using logarithm.
Problem 6 :
10^{x} = 100
Problem 7 :
3^{x} = ¹⁄₂₇
Problem 8 :
5^{3x - 1} = ¹⁄₆₂₅
Problem 9 :
2^{x} - 1 = 5
Problem 10 :
5^{x - 1 }- 2^{x} = 0
1. Answer :
3^{x} = 27
3^{x} = 3 ⋅ 3 ⋅ 3
3^{x} = 3^{3}
Equate the exponents.
x = 3
2. Answer :
2^{x - 5} = 2^{-4}
Equate the exponents.
x - 5 = -4
Add 5 to both sides.
x = 1
3. Answer :
7 ⋅ 3^{x} = 63
Divide both sides by 7.
3^{x} = 9
3^{x} = 3^{2}
Equate the exponents.
x = 2
4. Answer :
4^{x }= 7(2^{x}) + 18
(2^{2})^{x }= 7(2^{x}) + 18
(2^{x})^{2 }= 7(2^{x}) + 18
(2^{x})^{2 }- 7(2^{x}) - 18 = 0
Let a = 2^{x}.
a^{2 }- 7a - 18 = 0
(a - 9)(a + 2) = 0
a - 9 = 0 or a + 2 = 0
a - 9 = 0 a = 9 a^{ }= 3^{2} 3^{x }= 3^{2} x = 2 |
a + 2 = 0 a = -2 a = -2 3^{x }= -2 |
In 3^{x}, whatever real value (positive or negative or zero) we substitute for x, 3^{x} can never be negative. So we can ignore the equation 3^{x} = -2.
Therefore,
x = 2
5. Answer :
5^{x }= (√25)^{-7 }⋅ (√5)^{-5}
5^{x }= 5^{-7 }⋅ (5^{1/2})^{-5}
5^{x }= 5^{-7 }⋅ 5^{-5/2}
5^{x }= 5^{-7 - }^{5/2}
5^{x }= 5^{-19/2}
Equate the exponents.
x = -19/2
6. Answer :
10^{x} = 100
The above equation is in exponential form. Convert it to logarithmic form to solve for x.
x = log_{10}100
x = log_{10}100
x = log_{10}(10^{2})
x = 2log_{10}10
x = 2(1)
x = 2
7. Answer :
3^{x} = ¹⁄₂₇
The above equation is in exponential form. Convert it to logarithmic form to solve for x.
x = log_{3}(¹⁄₂₇)
x = log_{3}1 - log_{3}27
x = 0 - log_{3}(3^{3})
x = -3log_{3}3
x = -3(1)
x = -3
8. Answer :
5^{3x - 1} = ¹⁄₆₂₅
The above equation is in exponential form. Convert it to logarithmic form to solve for x.
3x - 1 = log_{5}(¹⁄₆₂₅)
3x - 1 = log_{5}1 - log_{5}625
3x - 1 = 0 - log_{5}(5^{4})
3x - 1 = -4log_{5}5
3x - 1 = -4(1)
3x - 1 = -4
Add 1 to both sides.
3x = -3
Divide both sides by 3.
x = -1
9. Answer :
2^{x} - 1 = 5
Add 1 to both sides.
2^{x} = 5
On the left side of the equation above, the base is 2. On the right side, 6 is not a power of 2. So, we can not get the same base on both sides.
Take logarithm on both sides and solve for x.
log(2^{x}) = log6
Using power rule of logarithm,
xlog2 = log6
Divide both sides by log2.
10. Answer :
5^{x - 1 }- 2^{x} = 0
Add 2^{x} to both sides.
5^{x - 1 }= 2^{x}
In the equation above, the base on the right side and the base left side are not same. And also, we can not make the base same on both sides using the rules of exponents.
Take logarithm on both sides and solve for x.
log(5^{x - 1}) = log(2^{x})
Using power rule of logarithm,
(x - 1)log5^{ }= xlog2
Using distributive property,
xlog5 - log5^{ }= xlog2
Subtract xlog2 from both sides.
xlog5 - log5^{ }- xlog2 = 0
Add log5 to both sides.
xlog5^{ }- xlog2 = log5
Factor.
x(log5^{ }- log)2 = log5
Divide both sides by (log5 - log2).
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