Question

5^4x-1=5^x+2
show me how to solve it pleaseeeee

Answer 1

\[5^{4x+1}=5^{x+2}\]
that show's it better

Answer 2

5^(4x)-5^x-1-2=0
5^(4x)-5^x-3=0
let u=5^x
so u^4=5^(4x)
so we have
u^4-u-3=0
lets see if we can use synthetic division to find any rational zeros if there are any
we need to look at plus or minus 1 and plus or minus 3
so lets try -1
-1| 1 0 0 -1 -3
| -1 1 -1 2
------------
1 -1 1 -2| -1

lets try 1
1 | 1 0 0 -1 -3
| 1 1 1 0
_______________
1 1 1 0 | -3
so neither 1 or -1 works
lets try 3
3| 1 0 0 -1 -3
| 3 9 27 78
_______________
1 3 9 26| 75
3 doesn't work
how about -3
-3| 1 0 0 -1 -3
| -3 9 -27 84
____________________
-3 -3 9 -28| 81
-3 doesn't work
so there are no rational solutions
we can use newton's method
do you know newton's method
but this method will only give us approximate answers

Answer 3

u=u0-f(u0)/f'(u0)
you know this method?

Answer 4

which problem are we doing the first or second problem you posted?

Answer 5

One sec let me read all that I'm totally lost :o

Answer 6

The answer to the 5^4x+1=5^x+2 = 1

Answer 7

\[5^{4x}-1=5^{x}+2\]
i was doing the first problem you posted

Answer 8

answer is, x=1

Answer 9

no there are no rational zeros

Answer 10

Oh okay I put that in wrong it's supposed to be \[5^{4x+1}=5^{x+2}\]

Answer 11

the answer to that is x=1 But how do you get that?

Answer 12

4x+1=x+2
4x-x=2-1
3x=1
x=1/3

Answer 13

it isn't 1
it is 1/3

Answer 14

maybe the teacher put it in wrong than thank youu

Answer 15

I have more if you don't mind helping

Answer 16

\[5^{4x-1}=5^{x+2}\] ?

Answer 17

4x-1=x+2
4x-x=2+1
3x=3
x=1

Answer 18

thank youu makes sense.

Answer 19

\[9^{x+1}=27^{x-1}\]

Answer 20

we don't have same base on both sides
but 9=3^2 and 27=3^3
so we have
\[3^{2(x+1)}=3^{3(x-1)}\]

Answer 21

since we have the same base we need the exponents to be the same
so we have
2(x+1)=3(x-1)

Answer 22

2x+2=3x-3
2x-3x=-3-2
-x=-5
x=5

Answer 23

that is right! yay!

Answer 24

how about this one i'm stuck i worked some of it out

Answer 25

\[3x-4 / 4-3x + 6-x / x-6 = 0\] I know the domain is 6, 4/3 the answer is no solution but how do you get that what I did was get like denominators which is (x-6)(4-3x) basically after doing all that i ended up with \[6x ^{4}-44x ^{2}+48=0\]

Answer 26

on top of (x-6)(4-3x)

Answer 27

\[\frac{3x-4}{4-3x}+\frac{6-x}{x-6}=0\] is this right?

Answer 28

yesssss

Answer 29

so then \[3x ^{2}-18x-4x+24+24-18x-4x+3x ^{?}\]

Answer 30

\[3x ^{2}\] *

Answer 31

ok so we know we cannot have zero on the bottom
so we don't want 4-3x=0 and we don't want x-6=0
so we don't want x=4/3 and we don't want x=6
\[\frac{3x-4}{(-1)(3x-4)}+\frac{(-1)(x-6)}{x-6}=0\]
\[\frac{1}{(-1)(1)}+\frac{(-1)(1)}{1}=0\]
-1+-1=0
-2=0 which is never true
so there are no solutions
i see the way you are going about it that is longer lol

Answer 32

Ah I'll go back to that one I have this huge final tomorrow at 7am! & All this stuff will be on it :(

Answer 33

did it not make sense what i have above?
you can go about it your way, but i think the way i above is much shorter?
here is what you were attempting on this attachement
let me know if there are any questions

Answer 34

Oh yeah that helps! That's what I was doing. thanksss :d

Answer 35

(b+5)(b+10)=6


x= -11, x= -4

How do you do that? :O

Answer 36

b^2+10b+5b+50=6
b^2+15b+50-6=0
b^2+15b+44=0
(b+4)(b+11)=0
b=-4
b=-11

Answer 37

wait you wrote it like this \[b ^{2+10b}\].. ?

Answer 38

(b+5)(b+10)=6^2 + (5+10)b+(5)(10) ?

Answer 39

i multiplyied b+5 and b+10

Answer 40

like foil?

Answer 41

yes