please help me with math

[geneticfreak]

okay, so i got this nasty equation.

(5x)^log5 - (3x)^log3 = 0

the answers are 1/15 and 0, but how do I get this, algebraically?
can anyone shed some light on this?

 

[geneticfreak]

somebody ?

 

[Instagib]

Bump for ya.

 

[Mister T]

x= 1/15

 

[BuckleDownBen]

Bump, because I tried it but couldn't solve it.

 

[Instagib]

Bump for a good math question

 

[Alphathree33]

(5x)^log5 - (3x)^log3 = 0

(5x)^log5 = (3x)^log3

log [ (5x)^log5 ] = log [ (3x)^log3 ]

log 5 log 5x = log 3 log 3x (bring exponents down with logarithm properties)

log 5 [ log 5 + log x ] = log 3 [ log 3 + log x ]

(log 5)^2 + log 5 log x = (log 3)^2 + log 3 log x

log 5 log x - log 3 log x = (log 3)^2 - (log 5)^2

[log 5 - log 3] log x = (log 3 - log 5)(log 3 + log 5)

log x = - (log 5 - log 3)(log 3 + log 5) / (log 5 - log 3)

log x = - (log 3 + log 5)

log x = - (log 15)

x = 1/15

Easy. ;-)

 

[bizmark]

ah very good Alphathree.

Regarding the 0 answer, (any constant)*x^(anything)=(any constant)*x^(anything else) always has 0 as an answer.

 

[Alphathree33]

IGNORE

 

[bmd]

The x=0 i can get, but i'm not sure how to solve for the 1/15. You can obviously graph it and see where it intersects with the X-axis, but that really isn't solving algebraically.

(5x)^(.69897)=(3x)^(.47712) (.69897 ~ log(5) and .47712 ~ log(3))
(0)^(.69897)=(0)^(.47712)
0=0

hmm...

OK i just got it.

(5x)^(.69897)=((3x)^(.47712) )*1
divide both sides by the right side
((5)^.69897)/((3)^(.47712))*x^(.221848745)=1
1.823536*x^(.221848745)=1
x^(.221848745)=1/1.823536
x^(.221848745)=.548385
raise both sides to the (1/.221848745) power
x=.066666 ~ 1/15

Edit: well done up above... just lost time typing this out =P.

 

[Alphathree33]

What I really wanted to say was if you sub zero for x, you get 0 - 0 = 0 which is true. not 1-1. That would be if x were in the exponent.

 

[Alphathree33]

Originally posted by: bmd
The x=0 i can get, but i'm not sure how to solve for the 1/15. [/i]

Way to not read my post.

 

[bmd]

Way to not read my post.

Sorry, but i loaded the page before you posted and spent time messing around typing my solution out. Way to be rude.

Edit: also, as you can see, I later did solve for the 1/15 in that same post. I didn't when i initially started typing my reply is all. Way to not read my post.

 

[Alphathree33]

Originally posted by: bizmark
ah very good Alphathree.

Regarding the 0 answer, (any constant)*x^(anything)=(any constant)*x^(anything else) always has 0 as an answer.

Those aren't constants, they are coefficients. Most polynomial functions lacking constants (i.e. "2", by itself, no variable attached) have zero as a root. So x^2 + x has a zero root, but x^2 + x + 3 doesn't have a zero root. (It doesn't have any real roots now that I look at it)

 

[BuckleDownBen]

Originally posted by: Alphathree33
(5x)^log5 - (3x)^log3 = 0

(5x)^log5 = (3x)^log3

log [ (5x)^log5 ] = log [ (3x)^log3 ]

log 5 log 5x = log 3 log 3x (bring exponents down with logarithm properties)

log 5 [ log 5 + log x ] = log 3 [ log 3 + log x ]

(log 5)^2 + log 5 log x = (log 3)^2 + log 3 log x

log 5 log x - log 3 log x = (log 3)^2 - (log 5)^2

[log 5 - log 3] log x = (log 3 - log 5)(log 3 + log 5)

log x = - (log 5 - log 3)(log 3 + log 5) / (log 5 - log 3)

log x = - (log 3 + log 5)

log x = - (log 15)

x = 1/15

Easy. ;-)

I managed the fist 6 or 7 steps. I do have one thing to add though. In the second step, where he takes the log of both sides, you can't take the log of 0, so you have to assume that 5x^log5 is different from 0. If 5x^log5 = 0 then 5x = 0 or x = 0, which you can verify is a solution.

 

[The Wildcard]

Hahaa, damm I just solved it too. Just some complex algebra. My solution was similar to that of bmd.
This solution might be simpler...

(5x)^log5 - (3x)^log3 = 0 (Original problem)
5^log5 * x^log5 - 3^log3 * x^log3 =0 (Just split up the exponent)
5^log5 * x^log5 = 3^log3 * x^log3 (Move it over to the right side)
5^log5 / 3^log3 = x^log5 / x^log3 (Collect variables to one side, and constants to the other side)
1.82353614167 = x^(log 5 - log 3) (Calculated the value of the left side and also simplfied the right side)
1.82353614167 ^ (1/(log5 - log3)) = x (Raise the left side to the inverse power of log5-log3) to isolate x
.066666666667 = x
1/15 =x

 

[bizmark]

Originally posted by: Alphathree33

Those aren't constants, they are coefficients. Most polynomial functions lacking constants (i.e. "2", by itself, no variable attached) have zero as a root. So x^2 + x has a zero root, but x^2 + x + 3 doesn't have a zero root. (It doesn't have any real roots now that I look at it)

Yeah, I meant coefficients. Constants was just the first word that came to mind. What I meant was clear, I think. The "constant" was clearly in the place of a coeffiecient.