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Solve the logarithm equation:log (2x-2) = log (x-2) + 1

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colinhartmann | Student, Undergraduate | eNotes Newbie

Posted April 14, 2011 at 3:38 AM via web

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Solve the logarithm equation:
log (2x-2) = log (x-2) + 1

Tagged with logarithm equation, math

2 Answers | Add Yours

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted April 14, 2011 at 3:41 AM (Answer #1)

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log (2x-2) = log (x-2) + 1

we will use logarithm properties to find the value of x.

First we know that log 10 = 1

==> log (2x-2) = log (x-2) + log 10

Now we know that log a + log b = log ab

==> log (2x-2) = log (10*(x-2)

==> log (2x-2) = log (10x -20)

Now, we know that if log a = log b ==> a = b

==> (2x-2) = 10x -20

==> 8x = 18

==> x = 18/8= 9/4= 2.25

Then the solution to the equation is 9/4 = 2.25.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted April 14, 2011 at 3:43 AM (Answer #2)

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We have log (2x-2) = log (x-2) + 1 and we have to solve for x.

log (2x-2) = log (x-2) + 1

=> log (2x-2) - log (x-2) = 1

use the property log a - log b = log(a/b)

=> log [(2x - 2)/(x - 2)] = 1

=> [(2x - 2)/(x - 2)] = 10

=> 2x - 2 = 10x - 20

=> 8x = 18

=> x = 18/8

=> x = 2.25

The solution of the equation is x = 2.25

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