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What is the answer for question 6. b) ? http://postimg.org/image/xusvq33e3/
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Using the logarithm rule of `log_(b)a+log_(b)c=log_(b)ac`
The value of x cannot be 0 because log 0 is undefined.
Therefore, x=3 is the required answer.
Posted by llltkl on May 17, 2013 at 1:37 PM (Answer #1)
You may move all the logarithms to one side, such that:
`log_7 x + log_7 (x - 1) - log_7 2x = 0`
Since the logarithms both sides share the same base` 7` , you may use the following logarithmic identities, such that:
`log x + log y = log(xy)`
`log x - log y = log(x/y)`
Reasoning by analogy yields:
`log_7 x + log_7 (x - 1) = log_7 x(x - 1)`
`log_7 x(x - 1) - log_7 2x = log_7 x(x - 1)/(2x)`
Replacing `log_7 x(x - 1)/(2x)` for log_7 x + log_7 (x - 1) - log_7 2x yields:
`log_7 x(x - 1)/(2x) = 0 => x(x - 1)/(2x) = 7^0`
`x^2 - x = 2x => x^2 - 3x = 0 `
Factoring out `x` yields:
`x(x - 3) = 0`
Using zero product property yields:
`x = 0`
`x - 3 = 0= > x = 3`
Testing the values `x = 0` and `x = 3 ` in equation yields:
`log_7 0 + log_7 (0 - 1) - log_7 2*0` invalid
`log_7 3 + log_7 (3 - 1) - log_7 2*3 = 0`
`log_7 6 - log_7 6 = 0` valid
Hence, evaluating the solution to the given logarithmic equation, yields `x = 3.`
Posted by sciencesolve on May 17, 2013 at 6:31 PM (Answer #2)
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