Question

I need help on how to solve this problem and how to set it up. 3^(3x-1)=9(x-3)

Answer 1

Typo on last post it should be: 3^(3x-1)=9^(x-3)

Answer 2

first u note that 9^(x-3) = 3^[2(x-3)]

therefore;
3x-1=2(x-3). solve this to get x.

Answer 3

where did the two come from?

Answer 4

thats because 3^2=9

Answer 5

as example, 9^2=3^(2*2)=81.
thats the same i done above. but instead of integer, the power in the question is an equation.

Answer 6

ok so if i have a problem like 3^(4x-2)=27^(x) then it would be 4x-2=3(x)??

Answer 7

yup. u got it right.

Answer 8

now what if i got a^(x-4)=[a^6][a^(3x-6)]

Answer 9

let say u hav (2^3)*(2^5), this wil be equal to 2^(3+5)

so in ur case, [a^6][a^(3x-6)]=a^[6+(3x-6)]

Answer 10

so then you would do a^x-4=a^[6+(3x-6)]
and then what do you do next?

Answer 11

yup.then do same as before, equate the power to get the x.
x-4=6+(3x-6)

Answer 12

alright now if you have a problem 9^(2x+1)=27^(x+2) then you have 2x+1=???

Answer 13

in this case, first u need to make the "base" equal. note that 9=3^2 and 27=3^3.

therefore,the eqn become: 3^[2(2x+1)]=3^[3(x+2)]

then equate the power as usual: 2(2x+1)=3(x+2)

Answer 14

ok so now i have 5^(2x+3)=(square root of 5)^(x+4)

Answer 15

u can write square root of 5 as 5^(1/2)

then u can proceed with 2x+3= (1/2)*(x+4)

Answer 16

why do you write it as 1/2? how did you get that

Answer 17

dont have to calculate anything for that.

its basic rule how to write it a square root.
\[\sqrt[m]{A} = A ^{1/m}\]

Answer 18

ohh ok haha also, i have (2a)^x=(4a^2)^x+2 and so would it be x=2(x+2)?

Answer 19

yes.thats right.

Answer 20

if the ratio of snicker to m&m's is 5 to 6 then the store could have

Answer 21

help

Answer 22

and on the last problem with the 1/2... what would the answer be?

Answer 23

i think it is x=-2/3. u got the same?

Answer 24

can u help me on my pro

Answer 25

no i got something different :/ umm i guess i did it wrong because i think you are right

Answer 26

try check it again. maybe a careless mistake.

Answer 27

ok now i got it :) okay well to switch gears a bit.. now i have to do log(3/5)(25/9)=x+4
**3/5 is the little bottom number**

Answer 28

remember that if \[\log_{a} b = c \]
this implies\[a ^{c}=b\]
so, u hav to change the initial problem into the 2nd form.
and u also hav to note that 25/9= (3/5)^-2

Answer 29

you mean 5/3^-2?

Answer 30

no. 25/9= (5/3)^2 = (3/5)^-2

Answer 31

ohh ok! im sorry to be asking so many questions its just that im so confused! ok on 2^x * 4^(x+5)=4^(2x-1) what do you do next?

Answer 32

since 4=2^2, then the eqn becomes;
2^x * 2^[2(x+5)]=2^[2(2x-1)]
this implies;
2^[x+2(x+5)]=2^[2(2x-1)]

therefore, x+2(x+5)=2(2x-1)

Answer 33

ohh ok umm what did you get on the previous problem? i dont know where x is supposed to go when you solve. i got this: (3/5)^4=(3/5)^-2

Answer 34

it should be (3/5)^(x+4)=(3/5)^-2

x included in "c" based on the general form i gave u before.

Answer 35

im going to bed.sorry cant help more.good luck with the problems..

Answer 36

ohh i forgot the x! so it should be -6.

Answer 37

ok thanks for all your help!!!! :)