Solving Logarithmic Equation with TI-89 Titanium: Step-by-Step Guide

[hostergaard]

Homework Statement

http://img205.imageshack.us/img205/2327/nummer1.jpg

Homework Equations

If you use ti-89 titanium, you write log with base(b) (c=number) as log(c,b) fx; the log of 3^x in base 5 is written log(3^x,5) (log₅(3^x))took me forever to figure out

The Attempt at a Solution

I filled out about 5 handwriten papers with attempts, all which failled. don't force me to write all my attempts since i find this to be very tedius.
simplified, the equation is;
10*5^x=3*5^-x+1

i put log₅ on each side geting:

log₅(5^x+3)-x=x+log₅(10)

then moving it about a bit i get:

2x-log₅(5^x+3)=log₅(10)

But here i get stuck...

 

[phymatast]

Try to Substitute: 5^x = y (with some restrictions on y)

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

Good luck

 

[hostergaard]

uh? sorry i couldt not get that to work, one of those 5^x 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.

 

[hostergaard]

Is there any way that i can get 3 out the log (or in a seperate) in of 2x-log₅(5^x+3)

 

[Gib Z]

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

$$a 5^{2x} + b 5^x + c$$.

Since 5^(2x) = (5^x)^2, its just a quadratic equation, like phymatast said.

 

[hostergaard]

i see, then i get 10*(5^x)²-5^x-3=0

then:

5^x=-(1$$\pm\sqrt{}$$-1²-4*10*(-3))/2*10

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

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

then i take log₅(5^x)=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]

Check your working in the quadratic formula, it should be -1, not 1 !

 

[HallsofIvy]

You shouldn't need the quadratic formula to solve $10y^2- y- 3= (2y+ 1)(5y- 3)$: which does NOT give y= 1/2, y= -3/5.

 

[phymatast]

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

 

[hostergaard]

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) which is the same as x=1+log₅(3)

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

 

[HallsofIvy]

-1/2 is impossible because the original equation would then involve log₅(-1/2) and logarithm is defined for negative values of x.

 

[hostergaard]

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.