Solve each of the following equations without using calculator.
Problem 1 :
ln x = -1
Solution :
ln x = -1
We know that the base of a natural logarithm is e.
Since ln referes to natural logarithm, its base is e.
ln_{e} x = -1
Convert to exponential form.
x = e^{-1}
Problem 2 :
log (x + 3) - 2 = 0
Solution :
log (x + 3) - 2 = 0
Add 2 to both sides.
log (x + 3) = 2
Since the base is not mentioned for log (x + 3), we have to understand that it is a common logarithm. We know that the base of a common logarithm is 10.
Then, we have
log_{10} (x + 3) = 2
Convert to exponential form.
x + 3 = 10^{2}
x + 3 = 100
Subtract 3 from both sides.
x = 97
Problem 3 :
x = log_{16} 32
Solution :
x = log_{16} 32
x = log_{16} (16 ⋅ 2)
Using the product of rule of logarithms,
x = log_{16} 16 + log_{16} 2
x = 1 + log_{16} 2
Problem 4 :
2log_{4} x + 1 = 0
Solution :
2log_{4} x + 1 = 0
Subtract 1 from both sides.
2log_{4} x = -1
Divide both sides by 2.
Convert to exponential form.
Problem 5 :
Solution :
Multiply both sides byt 4.
log_{x} 256 = 4
Convert to exponential form.
256 = x^{4}
4^{4} = x^{4}
4 = x
Problem 6 :
3ln (5x) = 10
Solution :
3ln (5x) = 10
Divide both sides by 3.
Convert to exponential form.
Problem 7 :
log_{3} x + log_{3} (x - 8) = 2
Solution :
log_{3} x + log_{3} (x - 8) = 2
Using the product rule of logarithm,
log_{3} x(x - 8) = 2
log_{3} (x^{2} - 8x) = 2
Convert to exponential form.
x^{2} - 8x = 3^{2}
x^{2} - 8x = 9
x^{2} - 8x - 9 = 0
Solve by factoring.
x^{2} - 9x + x - 9 = 0
x(x - 9) + 1(x - 9) = 0
(x - 9)(x + 1) = 0
x - 9 = 0 or x + 1 = 0
x - 9 = 0 x = 9 |
x + 1 = 0 x = -1 |
If x = -1, the arguments of both the logarithms in the given equation become negative. Logarithm is defined only for positive values of its argument. So, we can ignore x = -1.
Therefore,
x = 9
Problem 8 :
log_{4} x - log_{4} (2x - 3) = 3
Solution :
log_{4} x - log_{4} (2x - 3) = 3
Using the quotient rule of logarithm,
Convert to exponential form.
x = 64(2x - 3)
x = 128x - 192
192 = 127x
Problem 9 :
3log x - log (x^{2}) = 2
Solution :
3log x - log (x^{2}) = 2
Use the properties of logarithms.
log (x^{3}) - log (x^{2}) = 2
log x = 2
Since the base is not mentioned for log x, we have to understand that it is a common logarithm. We know that the base of a common logarithm is 10.
Then, we have
log_{10} x = 2
Convert to exponential form.
x = 10^{2}
x = 100
Problem 10 :
2log_{6} x + log_{6} (x + 1) = log_{6} (2x)
Solution :
2log_{6} x + log_{6} (x + 1) = log_{6} (2x)
Use the properties of logarithms.
log_{6} (x^{2}) + log_{6} (x + 1) = log_{6} (2x)
log_{6} x^{2}(x + 1) = log_{6} (2x)
Since two logarithms are equal with the same base (6), arguments can be equated.
x^{2}(x + 1) = 2x
x^{3} + x^{2} = 2x
x^{3} + x^{2} - 2x = 0
x(x^{2} + x - 2) = 0
x = 0 or x^{2} + x - 2 = 0
Solve x^{2} + x - 2 = 0 by factoring.
x^{2} + 2x - x - 2 = 0
x(x + 2) - 1(x + 2) = 0
(x + 2)(x - 1) = 0
x + 2 = 0 or x - 1 = 0
x + 2 = 0 x = -2 |
x - 1 = 0 x = 1 |
Solving the given equation, we get three values for x.
x = 0, -2, 1
If x = 0 or -2, the arguments of the logarithms in the given equation become zero or negative. Logarithm is defined only for positive values of its argument. So, we can ignore x = 0 and x = -2.
Therefore,
x = 1
Problem 11 :
Solution :
x(x - 6) = 5(x - 2)
x^{2} - 6x = 5x - 10
x^{2} - 11x + 10 = 0
x^{2} - 10x - x + 10 = 0
x(x - 10) - 1(x - 10) = 0
(x - 10)(x - 1) = 0
x - 10 = 0 or x - 1 = 0
x - 10 = 0 x = 10 |
x - 1 = 0 x = 1 |
Solving the given equation, we get three values for x.
x = 0, -2, 1
If x = 0 or -2, the arguments of the logarithms in the given equation become zero or negative. Logarithm is defined only for positive values of its argument. So, we can ignore x = 0 and x = -2.
Therefore,
x = 1
If x = 1, the arguments of both the logarithms on the left side of the given equation become negative. Logarithm is defined only for positive values of its argument. So, we can ignore x = 1.
Therefore,
x = 10
Problem 12 :
log (log x) = 1
Solution :
log (log x) = 1
Since the base is not mentioned for the logarithms in the above equation, we have to understand that they are common logarithms. We know that the base of a common logarithm is 10.
Then, we have
log_{10} (log_{10} x) = 1
log_{10} x = 10^{1}
log_{10} x = 10
x = 10^{10}
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