Question

(a^x + 1)/( b^x-1) = 10; what is x?

Answer 1

what are \(a\) and \(b\)

Answer 2

There are three variables how one can find three with just one equation??

Answer 3

i think a and b are constants
if so then hard to get x

Answer 4

are they integers

Answer 5

@radithya

Answer 6

Somehow I got:

\(a = 2, b = \sqrt{3}, x = 2 \)

Answer 7

i thought the solution might be \(x=x(a,b)\)
how could you solve three variables from on equation @waterineyes ?

Answer 8

= (x+1) log a - (x-1) log b = log 10

Answer 9

i think its
\[\huge \frac{a^x+1}{b^x-1}=10\]

Answer 10

\[\frac{a^x+1}{b^x-1}=\frac{10}{1}\]\[(a^x+1)=10(b^x-1)\[a^x+1=10b^x-1\]

Answer 11

@waterineyes 5/2≠10

Answer 12

Sorry 20/2 is to be taken...

Answer 13

\[(a^x+1)=10(b^x−1)\]\[a^x+1=10b^x-10\]\[a^x=10b^x-11\]

Answer 14

ya! thats wht i wants to say

Answer 15

\[\frac{a^x + 1}{b^x - 1} = \frac{50}{5}\implies a^x = 49 \implies a = 7 \quad and \quad x = 2\]

So, \(b = \sqrt{6}\)

\(a= 7, b = \sqrt{6}, x = 2\)

And It is also true that solutions can be many....

Answer 16

\[x=\log_a\left(10b^x-11\right)\]

Answer 17

There is x in the right hand side too..
It is not right...

Answer 18

For calculating x, it will be like:

x = something

And this something must not contain x...

Answer 19

You are going wrong @jiteshmeghwal9
You have calculated one variable from one equation and are going to substitute that variable in the same equation...

Answer 20

guys i think its impossible to calculate x in the form x=x(a,b) !!!!!! :D

Answer 21

Now you will get 0 @jiteshmeghwal9

Answer 22

ohhh ! yeah

Answer 23

im sure****

Answer 24

There will come \(10b^x - 10\) in the numerator @jiteshmeghwal9

Answer 25

infinite lines of solutions if a,b<7

Answer 26

thanks, its ok even if i don't get the answer