is this logarithm undefined?

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April 7th 2010, 08:24 PM
way123

is this logarithm undefined?

is this logarithm undefined where

log 8 -4 + log 8 -2 = 1

i know that you cannot take the log of a negiative but if you combine the logarithms to
log 8 (-4x-2) = 1
log 8 8 = 1
1=1
therefore it satisifies.

is that an acceptable answer or is it incorrect?
if so why is it not allowed?
April 7th 2010, 08:35 PM
Amer

Quote:

Originally Posted by way123

is this logarithm undefined where

log 8 -4 + log 8 -2 = 1

i know that you cannot take the log of a negiative but if you combine the logarithms to
log 8 (-4x-2) = 1
log 8 8 = 1
1=1
therefore it satisifies.

is that an acceptable answer or is it incorrect?
if so why is it not allowed?

log is defined for negative numbers but ln is not defined for negative numbers if the base is positive it is not defined for negative but for negatives it is ok like this

$log_{-2} -8 = 3$ since $(-2)^3 = -8$

the question is wrong since the base is positive

$log_{8} -4 = x \Rightarrow 8^x = -4$ but $8^x$ is always positive for all real numbers

so the question is wrong
April 7th 2010, 08:49 PM
Bacterius

Logarithmic functions are defined for positive numbers in the standard way and for negative numbers with their analytic continuation (involves complex numbers). So for the real numbers, logarithmic functions are only defined for positive numbers.

But I honestly don't understand your question. Could you express it in a "how can I solve this function for $x$ " way ?
April 7th 2010, 09:46 PM
way123

sorry that wasnt the original question i just put what i was not sure about. ill write the question and just look at your answer.

where i ran in to trouble was in in the check.

solve and check

log 8 (2-x) + log 8 (4-x) = 1
April 7th 2010, 09:51 PM
Bacterius

Quote:

Originally Posted by way123

sorry that wasnt the original question i just put what i was not sure about. ill write the question and just look at your answer.

where i ran in to trouble was in in the check.

solve and check

log 8 (2-x) + log 8 (4-x) = 1

You can use the property that $\log_b{(xy)} = \log_b{(x)} + \log_b{(y)}$ , then it is pretty straightforward :)

And to check, just plug in your result for $x$ and see if the equation holds.

Note : $\log_8{(a)} = \frac{\log{(a)}}{\log{(8)}}$

Where $\log$ is the standard logarithm key on your calculator :)
April 7th 2010, 09:57 PM
way123

no i know that.
i did that and got x ^2 -6x
therefore x is 0 and 6
and we have to check with left side, right side charts
so when i did the check for six i got

L.S. |R.S.
=log 8 -4 + log 8 -2 |=1

and then what do i do :S
April 7th 2010, 10:21 PM
Bacterius

To solve

$\log_8{(2-x)} + \log_8{(4-x)} = 1$

$\log_8{\left [ (2-x)(4-x) \right ]} = 1$

$8^{\log_8{\left [ (2-x)(4-x) \right ]}} = 8^1$

$(2-x)(4-x)= 8$

$(2-x)(4-x) - 8= 0$

$8 - 2x - 4x + x^2 - 8= 0$

$x^2 - 6x= 0$

$x(x - 6)= 0$

$x = 0 \ \ \textrm{and} \ \ x = 6$

To check

$\log_8{(2-6)} + \log_8{(4-6)} = 1$

$\log_8{(-4)} + \log_8{(-2)} = 1$

$\log_8{\left [ (-4)(-2) \right ]} = 1$

$\log_8{(8)} = 1$

$1 = 1$

Successful check !

$\log_8{(2)} + \log_8{(4)} = 1$

$\log_8{\left [ (2)(4) \right ]} = 1$

$\log_8{(8)} = 1$

$1 = 1$

Successful check !
______________________________________

Does it make sense now ? I do not know what you mean by left side/right side charts though. I guess the goal is to show that both sides are equal ...
April 7th 2010, 10:25 PM
Bacterius

Oh I think I know why you had a problem, you couldn't evaluate the logarithm of a negative number. Well, here you can combine both logs to obtain a positive number. If no combination is possible, then the answer is undefined and you have to conclude that the solution you found to the quadratic equation is outside the domain of the original function (involving logs).
April 8th 2010, 03:16 AM
HallsofIvy

Quote:

Originally Posted by Bacterius

To solve

$\log_8{(2-x)} + \log_8{(4-x)} = 1$

$\log_8{\left [ (2-x)(4-x) \right ]} = 1$

$8^{\log_8{\left [ (2-x)(4-x) \right ]}} = 8^1$

$(2-x)(4-x)= 8$

$(2-x)(4-x) - 8= 0$

$8 - 2x - 4x + x^2 - 8= 0$

$x^2 - 6x= 0$

$x(x - 6)= 0$

$x = 0 \ \ \textrm{and} \ \ x = 6$

To check

$\log_8{(2-6)} + \log_8{(4-6)} = 1$

$\log_8{(-4)} + \log_8{(-2)} = 1$

$\log_8{\left [ (-4)(-2) \right ]} = 1$

$\log_8{(8)} = 1$

$1 = 1$

Successful check !

No, not a successful check! The original problem was $\log_8{(2-x)} + \log_8{(4-x)} = 1$ , not $\log_8((2-x)(4-x))$
x= 0 satisfies the second of those but NOT the first equation. $log_8(2- 6)= log_8(-4)$ and $log_8(4- 6)= log_8(-2)$ do not exist so x= 6 does NOT satisfy the given equation. The fact that x= 6 satisfies $log_8((4-x)(2-x))$ is not relevant.

Quote:

$\log_8{(2)} + \log_8{(4)} = 1$

$\log_8{\left [ (2)(4) \right ]} = 1$

$\log_8{(8)} = 1$

$1 = 1$

Successful check !
______________________________________

Does it make sense now ? I do not know what you mean by left side/right side charts though. I guess the goal is to show that both sides are equal ...
April 8th 2010, 03:24 AM
Bacterius

Oh I didn't know that, thanks HallsOfIvy :)