If f(x)=log(15-x) (x+5) solve the equation f(x)=2 15-x is the base of log.
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We have the log to the base (15 - x)
f(x) = log(15-x) (x+5) = 2
=> (x + 5) = (15 - x)^2
=> x + 5 = 225 + x^2 - 30x
=> x^2 - 31x + 220 = 0
=> x^2 - 20 x - 11x + 220 =0
=> x( x - 20) - 11(x - 20) = 0
=> (x - 11)(x - 20) = 0
We get x = 11 and x = 20. Also, we can have logarithms with negative bases, so x can be 11 as well as 20.
Therefore the required values of x are 11 and 20.
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If f(x)=log(15-x) (x+5) solve the equation f(x)=2
15-x is the base of log.
Solution:
If f(x) = log (15-x) (x+5).
=> x+5 = (15-x)^ (f(x) by definition of log.
Therefore x+5 = (15-x) ^2.
x+5 = 225- 30x +x^2.
Subtract x+5:
x^2-31x+220 = 0.
(x-20)(x-11) = 0.
x- 20 = 0, or x-11 = 0.
So x= 20 , or x= 11 is the solution of f(x) = log (15-x) (x+5).
So x = 20, or x= 11.
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