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Solving a log proplem!

1. Homework Statement
http://img205.imageshack.us/img205/2327/nummer1.jpg [Broken]


2. Homework Equations
If you use ti-89 titanium, you write log with base(b) (c=number) as log(c,b) fx; the log of 3x in base 5 is writen log(3x,5) (log5(3x))took me forever to figure out
3. The Attempt at a Solution
I filled out about 5 handwriten papers with attempts, all wich failled. dont force me to write all my attempts since i find this to be very tedius.
simplified, the equation is;
10*5x=3*5-x+1

i put log5 on each side geting:

log5(5x+3)-x=x+log5(10)

then moving it about a bit i get:

2x-log5(5x+3)=log5(10)

But here i get stuck...
 
Last edited by a moderator:
Try to Substitute: 5^x = y (with some restrictions on y)

You'll end up solving a quadratic equation in y!

Good luck
 
uh? sorry i couldt not get that to work, one of those 5x has a negative sign... (i am not sure i did it right. where in the the proces shuld i put it in?)... sorry, i still need help.
 
Is there any way that i can get 3 out the log (or in a seperate) in of 2x-log5(5x+3)
 

Gib Z

Homework Helper
3,344
3
You should try phymatast's suggestion again. Multiply both sides by 5^x. Then see if you can pull out any factors and make the equation look like

[tex] a 5^{2x} + b 5^x + c[/tex].

Since 5^(2x) = (5^x)^2, its just a quadratic equation, like phymatast said.
 
i see, then i get 10*(5x)2-5x-3=0

then:

5x=-(1[tex]\pm\sqrt{}[/tex]-12-4*10*(-3))/2*10

that gives me that
5x=1/2
5x=-3/5

with 5x=-3/5 my calculator goes haywire so gues thats an imposiple answer (because of the -)

then i take log5(5x)=log(1/2)
wich gives me x=-log5(2)
wich gives me an asnwer that dosent fulfil that the anser shuld be given in a+log5b (exept if i say 0-log52 i quess)

Is that the correct answer tough? need a litlle bit more help here!
 

Gib Z

Homework Helper
3,344
3
Check your working in the quadratic formula, it should be -1, not 1 !
 

HallsofIvy

Science Advisor
Homework Helper
41,683
865
You shouldn't need the quadratic formula to solve [itex]10y^2- y- 3= (2y+ 1)(5y- 3)[/itex]: which does NOT give y= 1/2, y= -3/5.
 
it's 3/5 and not -3/5 and this will give you your answer using basic properties of logarithmic functions.

hint: after you get 5^x = 3/5

take log base 5 on both sides
 
aah, it seem i have made a mistake; the two possible anser is 3/5 and -1/2.
So, -1/2 is imposible (could someone explain why?). that leaves me with 3/5 using log5 on both side i get x =log5(3/5) using the rule that log(a/b)=log(a)+log(b) i get x=log5(5)+log5(3) wich is the same as x=1+log5(3)

Heruka, i finnaly got it! thanks guys, you are all great!:biggrin:
 

HallsofIvy

Science Advisor
Homework Helper
41,683
865
-1/2 is impossible because the original equation would then involve log5(-1/2) and logarithm is defined for negative values of x.
 
Rigth stupid me; its because log5(-1/2) Reads
"what power must 5 be raised to, to get -1/2" And it is imposible because you can't get a negative number by raising a positive number.
 

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