another log question...

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pinkcalculator

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Oct 13, 2009
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I know the rule for logs is that logb(x) -logb(y) = logb (x/y).
However, I'm not sure how that rule applies when there is a fraction subtracted...
Like in this:

log2 (x-(11-x/x-3) +3=4.
I know that you have to get the log alone
log2 (x-(11-x/x-3)=1
and that the negative must be factored out??? Maybe, but how??

This is as far as I've gotten, and I'm not sure what to do about the negative fraction.
Any help would be greatly appreciated!
 
Hard to tell what you really mean without additional grouping symbols.

If you mean \(\displaystyle \log_2(x-\frac{11-x}{x-3})+3=4\) you might use the definition of logs, namely...

log[sub:1ddjuc4n]b[/sub:1ddjuc4n] N = x implies that b[sup:1ddjuc4n]x[/sup:1ddjuc4n]=N.
 
pinkcalculator said:
I know the rule for logs is that logb(x) -logb(y) = logb (x/y).
However, I'm not sure how that rule applies when there is a fraction subtracted...
Like in this:

log2 (x-(11-x/x-3) +3=4.
I know that you have to get the log alone
log2 (x-(11-x/x-3)=1
and that the negative must be factored out??? Maybe, but how??

This is as far as I've gotten, and I'm not sure what to do about the negative fraction.
Any help would be greatly appreciated!
Click to expand...

The way you wrote it - applying PEMDAS - it translates to:

\(\displaystyle x - (11 - \frac{x}{x} - 3) = x - (11 - 1 - 3) = x - 7\)

If that is NOT what you meant - please fix your post.

If thais what you meant then - from definition of log function:

x - 7 = 2[sup:3ndv0j3x]1[/sup:3ndv0j3x]

x = 9
 
Loren said:
Hard to tell what you really mean without additional grouping symbols.

If you mean \(\displaystyle \log_2(x-\frac{11-x}{x-3})+3=4\) you might use the definition of logs, namely...

log[sub:3pcy8duw]b[/sub:3pcy8duw] N = x implies that b[sup:3pcy8duw]x[/sup:3pcy8duw]=N.
Click to expand...

That's how the problem looks.
I solved it like this:
log2(x- 11-x/x-3) = 1
2^1= x-11x/x-3)
2-x= -11-x/x-3
2-x(x-3) =-11-x
2x-6-x^2+3x=-11-x
-x^2 +5x-6=-11-x
Then I factored to find that x=-1, x=6.
 
pinkcalculator said:
Loren said:
Hard to tell what you really mean without additional grouping symbols.

If you mean \(\displaystyle \log_2(x-\frac{11-x}{x-3})+3=4\) you might use the definition of logs, namely...

log[sub:3zpwbqmm]b[/sub:3zpwbqmm] N = x implies that b[sup:3zpwbqmm]x[/sup:3zpwbqmm]=N.
Click to expand...

That's how the problem looks.
I solved it like this:
log2(x- 11-x/x-3) = 1
2^1= x-11x/x-3)
2-x= -11-x/x-3
2-x(x-3) =-11-x
2x-6-x^2+3x=-11-x
-x^2 +5x-6=-11-x
Then I factored to find that x=-1, x=6.
Click to expand...

To check correctnes of your answer - you need to substitute -1 & 6 respectively, in the given equation and see if you get back an identity.
 
An identity?? I checked that both made the (x-(11-x/x-3) greater than zero. Is that the same thing??
 
pinkcalculator said:
An identity?? I checked that both made the (x-(11-x/x-3) greater than zero. Is that the same thing??
Click to expand...
Sort of. But really you need to check as follows:

\(\displaystyle log_2\left (x - \frac{11-x}{x-3}\right ) + 3 = 4\)

\(\displaystyle log_2\left (-1 - \frac{11-(-1)}{(-1)-3}\right ) + 3 = 4\)

\(\displaystyle log_2\left (-1 - \frac{12}{(-4)}\right ) + 3 = 4\)

\(\displaystyle log_2\left (-1 - (-3)\right ) + 3 = 4\)

\(\displaystyle log_2\left (2\right ) + 3 = 4\)

\(\displaystyle 1 + 3 = 4\)

\(\displaystyle 4 = 4\)..........................Identity
 
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