log (x-9) + log x = 1

- justme

log (x-9) + log x = 1

- katieb

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- radar

\[10^{x-9}+10^{x}=10^{0}\]
x-9+x=0
2x=9
x=4.5
check:\[10^{4.5-9}+10^{4.5}=10^{0}=1\]

- justme

wow thanks! the way you wrote it made it so much easier to understand ( :

- amistre64

log(x^2 - 9x) = 1
x^2 - 9x = 10^1 = 10
x^2 - 9x - 10 = 0
(x-10)(x+1) = 0
x = 10, -1 ......
log(10-9) + log(10) = 1 ??
log(1) + 1 = 1
log(1) = 0
10^0 = 1 ... did i do that right?

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## More answers

- amistre64

log(-1-9) = log(-10)...-1 aint an option tho :)

- amistre64

log(4.5 - 9) + log(4.5) = 1
log(-4.5) + log(4.5)
logs cant be negative....

- amistre64

bases dont add if they aint got the same exponents
x^2 + x^3 doesnt equal x^5

- radar

Can not you let them be negative to solve this problem?? lol

- amistre64

well; we will have to write the board of logaritms and enter a plea, then the senate will ahve to vote on it .... its just alot of work for one litle problem :)

- justme

base is 10 because it is not given. so all the bases here are the same.

- amistre64

doesnt matter; bases dont add without like exponents
x^2 + x^5 DOESNT EQUAL x^7

- amistre64

10^a + 10^b DOESNT EQUAL 10^(a+b)

- amistre64

logs arent numbers; so they dont work like numbers; they are exponents and follow the rules of exponents

- radar

Yes, I do see the flaw in such an elegant solution, sorry about that justme, please ignore.

- justme

so x=4.5 right? seems to work when checked.

- amistre64

log(x-9) + log(x) = 1; exponents add when bases multiply
log((x)(x-9)) = 1
log(x^2 -9x) = 1
10^log... = 10^1
x^2 -9x = 10
x^2 -9x -10 = 0
x= 10, -1 ; -1 is illegal to use in the equation becasue log(-1-9) doesnt exist

- radar

adding as I did was really multiplying, and of course 10^x*10^-x would equal 1.

- amistre64

4.5 is illegal to use; becasue log(-4.9) does not exist

- amistre64

you can tell by graphing the log function and going over to -10 and -4.5 and looking around to find a y value.... it aint there :)

- radar

You have convinced me this problem is above my pay grade, please post the correct solution if you can or anyone can solve it.

- amistre64

the correct solution is:
x = 10.
And we all get lost in the math eventually ;)

- justme

how did you get that?

- amistre64

........ if you scroll up and read the post, i actually typed it out a few times lol

- radar

wouldn't that result in 11=1 we know thats not right.

- amistre64

log(10-9) + log(10) = 1
log(1) + 1 = 1
log(1) = 0
10^log(1) = 10^0
1 = 10^0
1 = 1

- amistre64

also known as:
0 + 1 = 1

- radar

Thanks

- radar

It will work.

- radar

Justme, did you see the flaw in the way I worked it? I hope I didn't mess you up. But the addition was illegal, as we were dealing with exponents, adding is exponents is the same as multiplying, and your problem was addition. I learned a lot on that thing!!!

- justme

radar, me too that was the best lesson on log I have seen. thanks!

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