log_{b}a = k
Since the base of the logarithm is b, exponentiate both sides with base b.
Solved each of the following logarithmic equations by exponentiating.
Problem 1 :
log_{4}(x - 5) = 3
Solution :
Since the base of the logarithm is 4, exponentiate both sides of the equation with base 4.
x - 5 = 4^{3}
x - 5 = 64
Add 5 to both sides.
x = 69
Problem 2 :
log(5x) = log(2x + 9)
Solution :
Since the base is not given for both the logarithms, we have to assume that they are common logarithms. Hence the base for both the logarithms is 10.
log_{10}(5x) = log_{10}(2x + 9)
Since the base of the logarithms is 10, exponentiate both sides of the equation with base 10.
5x = 2x + 9
Subtract 2x from both sides.
3x = 9
Divide both sides by 3.
x = 3
Problem 3 :
ln(5x - 1) = ln(7x - 9)
Solution :
Since the given logarithms are natural logarithms, their base is e.
So, exponentiate both sides of the equation with base e.
5x - 1 = 7x - 9
Subtract 7x from both sides.
-2x - 1 = -9
Add 1 to both sides.
-2x = -8
Divide both sides by (-2).
x = 4
Problem 4 :
Solution :
The bases of the logarithms are different, they are 3 and 9.
Here, 3 can not be written as a power of 9.
But 9 can be written as a power of 3, that is
9 = 3^{2}
So, exponentiate both sides of the equation with base 9.
3x + 7 = 2x + 10
Subtract 2x from both sides.
x + 7 = 10
Subtract 7 from both sides.
x = 3
Problem 5 :
log_{14}(x - 3) + log_{14}(x + 2) = 1
Solution :
log_{14}(x - 3) + log_{14}(x + 2) = 1
log_{14}(x - 3) + log_{14}(x + 2) = log_{14}14
log_{14}[(x - 3)(x + 2)] = log_{14}14
log_{14}(x^{2} - x - 6) = log_{14}14
Since the base of the logarithms is 14, exponentiate both sides of the equation with base 14.
x^{2} - x - 6 = 14
Subtract 14 from both sides.
x^{2} - x - 20 = 0
Solve by factoring.
x^{2} - 5x + 4x - 20 = 0
x(x - 5) + 4(x - 5) = 0
(x - 5)(x + 4) = 0
x - 5 = 0 x = 5 |
x + 4 = 0 x = -4 |
The aruguments of logarithms are always positive or greater than zero.
x - 3 > 0 x > 3 |
x + 2 > 0 x > -2 |
The common solution of the inequalities x > 3 and x > -2 is
x > 3
Since x > 3, the solution x = -4 can not be acccepted.
Therefore,
x = 5
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