Question

Solve for x (Logs)

Answer 1

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

Answer 2

(x+1) log 5 = (x-1) log 2
(x+1)/(x-1) = log 2/log 5
use compenendo dividendo :
x= log 10 / log (0.4)

Answer 3

was that helpful?

Answer 4

so you bring (x-1) to the left and logs to the right

Answer 5

log (a) is simply just another constant,,like 1,7 etc..

Answer 6

so how did you get from x+1/x-1 = log2/log5 to x = ?

Answer 7

just use cross multipication,,you'll get the same ans..
i used method called compenendo dividendo (not sure about the spellings)

Answer 8

so do i let log 5 = 0.6989 and log 2 = 0.3010?

Answer 9

for a/b = c/d
add 1 to both sides
(a+b)/b = (c+d)/d
and subtract 1
(a-b)/b = (c-d)/d
on dividing 2nd and 4th eqns..
we get (a+b)/(a-b) = (c+d)/(c-d)
this is what i used..

Answer 10

if the base is 10 then yes..

Answer 11

I've never done compenendo dividendo, and I'm still a little confused :(

Answer 12

forget that thing for now,,just use simple cross multipication,,you'll get the same ans..

Answer 13

so (x+1)log4 = (x+2)(log5)?

Answer 14

huh?? whats this?
its (x+1) log 5 = (x-1) log 2
note : log a^b = b log a

Answer 15

Sorry, yeah. Then?

Answer 16

the next step is my problem

Answer 17

open up brackets
you have
x log 5 +log 5 = x log 2 - log 2

Answer 18

now constants on one side,,variables on the other and solve for x..

Answer 19

xlog5-xlog2=-log2-log5?

Answer 20

yes..take x common from LHS ..rest should be easy..

Answer 21

k

Answer 22

thanks

Answer 23

happy to help ^^