Question

If x= z and y=a satisfys the equation 2^(x+y) = 6^y and 3^(x-1)= 2^(y+1) then z+a=?

Answer 1

options are \[\log_{3}6 \] \[\log_{3/2} 6\]

Answer 2

@satellite73 @Callisto @experimentX @badi

Answer 3

Take log on both sides

\[\large \log(2^{x+y}) = \log(6^y)\]

Applying power rule, we get

(x+y)log 2 = y log 6

Similarly apply for the 2nd equation and get two equations and try to solve

Answer 4

\[ 6 = 2^{log_2 6} and \; 3 = 2^{log_2 3}\]
do some algebric manipulation

Answer 5

Also log 6 = log(2*3) = log 2 + log 3

Answer 6

i am stuck @experimentX @satellite73 @Callisto plz help

Answer 7

just some algebraic manipulation ... cancel out the bases .... equate the powers
two eqautions ... two unkonwsn ... solve them!!

Answer 8

oh......no !!!!! getting an error plzz help

Answer 9

Yeah you should get : \[\log_{3/2}6 \]

Answer 10

the correct answer is that which @siddhantsharan wrote

Answer 11

\[ (x+y) = y\log_2 6 \]
\[ (x-1)\log_23 = (y+1)\]
try solving them

Answer 12

Dont solve them. Manupilate to get x+ y.

Answer 13

Take everything on a base 3 by taking log on both sides.
\[(x + y)\log_{3}2 = y \log_{3} 6\]

And x -1 = log 2(y +1)

Answer 14

From the first you get x log2 = y

Answer 15

Then Add the equations And solve for x + y

Answer 16

@shivam_bhalla http://openstudy.com/study?signup#/updates/4fc21268e4b0964abc83988e

Answer 17

@shivam_bhalla u visited my question.

Answer 18

@satellite73 @Mertsj

Answer 19

I added the equations like you said but couldn't make any progress from there.

Answer 20

Okay. Gimme a minute to type it. :)

Answer 21

Assume all logs as base 3 unless written otherwise.
\[y = xlog2 \]
\[x - 1 = ylog2 - \log2\]

Adding,

\[(x + y) - 1 = (\log2)(x+y)\]

Answer 22

Let x + y be t.

t - 1 = tlog2 => t = 1 / ( 1 - log2)