I need some math help

[Ed!]

Shameful Mathematics from people in this thread. You cannot divide by e^4x from both sides, one is a multiply, the other is an addition. Also e^7x-e^4x is not e^3x:

e^7 - e^4 = 1042,035....
e^3 = 20.0855....

Not the same yo

The solution I get is as followed:
e^(3x+4x) = e^(4x) - 15
e^7x - e^4x = -15

You can cannot do anything with logs at the moment, I think you have to use substitution: u=e^x
u^7 - u^4 +15 = 0.

Have fun with that, guarantee you'll end up with complex numbers and logs

 

[ShaRpY HaRdY]

So @Delik had it right last time he posted I think. I put him on the right path though! :dawg:

yeah he had it right the 2nd time, didnt even notice this got out to a second page that's why I missed it I think.

 

[Jonathan]

Shameful Mathematics from people in this thread. You cannot divide by e^4x from both sides, one is a multiply, the other is an addition. Also e^7x-e^4x is not e^3x:

e^7 - e^4 = 1042,035....
e^3 = 20.0855....

Not the same yo

The solution I get is as followed:
e^(3x+4x) = e^(4x) - 15
e^7x - e^4x = -15

You can cannot do anything with logs at the moment, I think you have to use substitution: u=e^x
u^7 - u^4 +15 = 0.

Have fun with that, guarantee you'll end up with complex numbers and logs

One of the first things I was taught was when dealing exponents, for example e^7 - e^4 you just take them away, to give e^3

 

[Ed!]

One of the first things I was taught was when dealing exponents, for example e^7 - e^4 you just take them away, to give e^3

But it isn't e^7 it is 7x, far as I know, you cannot subtract. I thought that is what everyone did when answering, best get out a rule and slap some hands for poor maths

 

[ShaRpY HaRdY]

On one side you have e^4x the other is (e^4)x turning it into e^4x you can subtract one from the other to cancel them out.. that leaves you with e^3x = e^-15 and at that point you don't worry about the e's because all you're doing is solving for x and since the bases are the same you just set the exponents equal to eachother.

 

[Trip in the Head]

On one side you have e^4x the other is (e^4)x turning it into e^4x you can subtract one from the other to cancel them out.. that leaves you with e^3x = e^-15 and at that point you don't worry about the e's.

Ah, so the answer is the same as what Jono said, he just technically went about doing it the wrong way (which is technically what I said). The two e^4x's cancel out independently from the e^3x. Right?

 

[ShaRpY HaRdY]

Ah, so the answer is the same as what Jono said, he just technically went about doing it the wrong way (which is technically what I said). The two e^4x's cancel out independently from the e^3x. Right?

yup, x = -5 is the correct answer.

^ Used some weird math site so I could show you how it works out without typing it even though Jono got it.

My bad you don't subtract it, you do divide it out because the e^4x-15 becomes e^4x times e^-15

 

[Trip in the Head]

yup, x = -5 is the correct answer.

^ Used some weird math site so I could show you how it works out without typing it even though Jono got it.

My bad you don't subtract it, you do divide it out because the e^4x-15 becomes e^4x times e^-15

Lucky Lucky @Delik if it had been e^2x * e^5x the math wouldn't have worked the same it seems

 

[ShaRpY HaRdY]

I got like a test on this tomorrow and I'm so burnt out lol.

 

[Trip in the Head]

It was also not evident to me that it was

in your picture. It almost looked like
(e^4x)-15 lol. That little bit of height makes a difference

 

[ShaRpY HaRdY]

It was also not evident to me that it was

in your picture. It almost looked like
(e^4x)-15 lol. That little bit of height makes a difference

Yeah I'm not sure if I said that earlier or if it was on a diff forum but to me the first time it straight looked like (e^4)^x and then -15 separate which is where I was first stuck initially.

 

[Nero_x3]

E ISNT A VARIABLE ITS A LOGARITHMIC FUNCTION.

 

[ShaRpY HaRdY]

E ISNT A VARIABLE ITS A LOGARITHMIC FUNCTION.

Not sure who you are referring to.. but you can divide a logarthimic function by a logarithmic function if the base is the same.
The answer isn't wrong and neither is the way I showed it, I got 100% on the assignment.

 

[Trip in the Head]

Not sure who you are referring to.. but you can divide a logarthimic function by a logarithmic function if the base is the same.
The answer isn't wrong and neither is the way I showed it, I got 100% on the assignment.

Exactly THANK YOU

 

[Nero_x3]

Exactly THANK YOU

Cept it's not right, and logarithms can only be subtracted if substituted correctly, which it wasn't.