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Solve the logarithmic equation algebraically:

0 votes

Solve the logarithmic equation algebraically:

 

asked Dec 22, 2017 in PRECALCULUS by anonymous
reshown Dec 22, 2017 by bradely

1 Answer

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(a)

log_2 (10 + 3x)   =   5

According to basic rule of logorithm

10 + 3x   =   2^5

10 + 3x   =   32

3x   =   32 - 10

3x   =   22

x  =  22/3

The solution is x = 22/3

 

(b)

log x + log (x - 9)   =   1

log [ x(x - 9) ]   =   1

log (x^2 - 9x)   =   1

According to basic rule of logorithm

(x^2 - 9x)   =   10^1

x^2 - 9x   =  10

x^2 - 9x - 10   =   0

x^2 + x - 10x - 10   =   0

x(x + 1) - 10(x + 1)   =   0

(x + 1)(x - 10)   =   0

x + 1   =   0       ;         x - 10   =  0

x   =  - 1            ;         x   =   10

The solutions are x = -1, 10.

answered Dec 22, 2017 by homeworkhelp Mentor

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