Question

PLEASE HELP
1. 2^x-1 = 5^2x+6

Answer 1

@zepdrix

Answer 2

is it \[2^{x*1}=5^{2x+6}\]?

Answer 3

yes :)

Answer 4

x-1 actually

Answer 5

ok be prepared to do some algebra

Answer 6

oh goodie

Answer 7

you gotta start with \[(x-1)\ln(2)=(2x+6)\ln(5)\] as a first step

Answer 8

don't be confused by this \(\ln(2)\) and \(\ln(5)\) are just some numbers

Answer 9

constants in other words, we are solving for \(x\) so the first thing you do is to distribute on both sides of the equal sign

Answer 10

i think i get it

Answer 11

we can take an additional step at the beginning, but it is unnecessary
the math teachers say "take the log" and write \[\ln(2^{x-1})=\ln(5^{2x+6})\] but then the next step is to rewrite that (the whole point is to bring the \(x\) out of the sky) and write \[(x-1)\ln(2)=(2x+6)\ln(5)\]

Answer 12

ready for the algebra?

Answer 13

oh i get it now. go on please :)

Answer 14

distribute first, remembering ( i repeat) that \(\ln(2)\) and \(\ln(5)\) are just numbers
so first second step is to write \[\ln(2)x-\ln(2)=2\ln(5)x+6\ln(5)\] let me know when you digest that

Answer 15

on the left side is it - ln(2) because in x - 1 the one has a - in front ?

Answer 16

it is the distributive property in action
\((b+c)a=ab+ac\)
here \(b=x.c=-1,a=\ln(2)\) so \[(x-1)\ln(2)=\ln(2)\times x-1\times \ln(2)\]

Answer 17

more succinctly written as \[\ln(2)x-\ln(2)\] same story on the other side

Answer 18

okay so what would the next step be?

Answer 19

you are solving for \(x\) so you have to get everything with an \(x\) on one side of the equal sign, everything else on the other

Answer 20

it doesn't matter which side you pick but lets put the \(x\) stuff on the left ok?

Answer 21

again don't be confused the the logs, it is like solving \[Ax-A=2Bx+6B\] for \(x\)

Answer 22

okay so i do the opposite operation right which would be subtracting 2ln(5)x ...?? or am i really off

Answer 23

no you are not off (not that i know of anyway) you are exactly right

Answer 24

okay phew

Answer 25

\[\ln(2)x-\ln(2)=2\ln(5)x+6\ln(5)\\
\ln(2)x-2\ln(5)x-\ln(2)=6\ln(5)\] after subtracting

Answer 26

yes i got that right!

Answer 27

now we got to get the term without the \(x\) on the other side

Answer 28

yay!

Answer 29

yay indeed!

Answer 30

so since you are on a roll, what do you do now to get the term without an x on the right?

Answer 31

so + in(2) on both sides

Answer 32

bingo

Answer 33

and from there...

Answer 34

\[\ln(2)x-2\ln(5)x=6\ln(5)+\ln(2)\] is where we should be

Answer 35

yes :)

Answer 36

ok now here is the only part that requires some sophistication
before we do it, could you solve \[Ax+Bx=C\] for \(x\) if you had to ?
"no" is a fine answer, just asking

Answer 37

for which side?

Answer 38

solve for \(x\)

Answer 39

if not, that is fine, i will show you

Answer 40

yes please

Answer 41

ok first imagine we had to solve \[3x+7x=15\] we would add first and get \[10x=15\] then divide by 10

Answer 42

yes i understand that

Answer 43

if we have \[Ax+Bx=C\] we can't really add (or combine like terms) but we would do the same thing \[(A+B)x=C\] so \[x=\frac{C}{A+B}\]

Answer 44

in other words, factor out the \(x\) so we know what to divide by
here we have \[\ln(2)x-2\ln(5)x=6\ln(5)+\ln(2)\] so we have to factor out the \(x\) on the left and get \[(\ln(2)-2\ln(5))x=6\ln(5)+\ln(2)\]

Answer 45

ooooh yes i see

Answer 46

let me know if that last step is clear, it is the only part of this that is confusing

Answer 47

i going to take that as a "i get it"
final step is to divide to finish with \[x=\frac{6\ln(5)+\ln(2)}{\ln(2)-2\ln(5)}\]

Answer 48

yes i understand

Answer 49

told you it was a bunch of algebra
the only not algebra part was taking the log at the beginning to get the variable on the ground floor

Answer 50

want to check it?

Answer 51

im doing the calculations now :)

Answer 52

here is an easy check
http://www.wolframalpha.com/input/?i=+2^%28x-1%29+%3D+5^%282x%2B6%29
you can see that it is right

Answer 53

now i have a question

Answer 54

what the monkey is a "wilder monday"?

Answer 55

haha i get that a lot

Answer 56

its actually my name. Wilder-Monday Turner is my full name

Answer 57

i have no idea why my mother did that but people just call me Monday

Answer 58

that is a pretty cool name
i know someone named "tuesday wolf" which i also like

Answer 59

thats a nice name!

Answer 60

and not that you asked, but "satellite73" is my car
http://www.dvap.com/salvage/1973-plymouth-satellite-73pl5238d/
mine is nicer but similar

Answer 61

i feel bad for not asking ahaha. thats a sick ride you got there

Answer 62

lol don't feel bad
you got more or are you done?

Answer 63

more questions? yes i have more if you would like to help me :)

Answer 64

sure why not? it is nice to help someone who actually participates too

Answer 65

i like being involved because it helps me understand better instead of having someone do it all for me

Answer 66

ill make a new question or do you want to keep going on this one?

Answer 67

you can go ahead an post here if you like

Answer 68

okay :)
1. 3 + e^3x+2 = 23

Answer 69

it is \[1.3+e^{3x+2}=23\]?

Answer 70

No no its just 3

Answer 71

sorry bout that

Answer 72

ooh \[3+e^{3x+2}=23\]

Answer 73

yes yes

Answer 74

this is way way (way) easier than the last one
a) subtract 3 from both sides

Answer 75

done :)

Answer 76

rewrite in equivalent logarthmic form, i.e go from \[e^{3x+2}=20\] to \[3x+2=\ln(20)\] that is the only none algebra part

Answer 77

"non algebra" part

Answer 78

then solve as normal, subtract 2, then divide by 3

Answer 79

thats it ?! easy!

Answer 80

way way (way) easier than the last one
you can just about do the last steps in your head
you should have \[x=\frac{\ln(20)-2}{3}\]

Answer 81

yep and i got 0.3319...

Answer 82

if you are supposed to answer as a decimal, then that is probably correct

Answer 83

yes decimals are fine

Answer 84

can you help with two more?

Answer 85

sure
btw your answer is right

Answer 86

okay thanks :)

Answer 87

these next ones look long and hard

Answer 88

and i just realized that sounded very wrong whoops

Answer 89

oh great (maybe they are not so long and hard lol)

Answer 90

maybe im over reacting lol

Answer 91

log (2-1) = log (4x-3) - log x

Answer 92

there must be a typo there right? is it really \[\ln(2-1)\] on the left?

Answer 93

it says log(2-1)

Answer 94

and then the rest on the other side

Answer 95

is that bad

Answer 96

ok if that is what it really says (seems weird) then it is pretty clear that \(2-1=1\)right?

Answer 97

yes that seems about right lol

Answer 98

so you really have \[\log (1) = \log (4x-3) - \log (x)\]