Question

How to solve for x ?
log 4^(x+3) + log 4^(2-x) = 1

Answer 1

use the identities : log(A.B) =log(A) + log(B) , and log(10)=1

Answer 2

wat is the base of log????

Answer 3

\[\log_{?} 4^(x+3+2-x)\]

Answer 4

\[\log_{4^{\left( x+3 \right)}} + \log_{4^{\left( 2-x \right)}} = 1\]

Answer 5

@r.gupta04 now check the Q mate

Answer 6

change the base

Answer 7

the one i mentioned was right...i:e its log to the base 10 by default ...and ln(x) is to the base e

Answer 8

make base of both log common

Answer 9

@r.gupta04 yeah sure u were right ...but he asked the Q which he dont know how to type

Answer 10

guys the base of log is 4

Answer 11

@gorv @r.gupta04

Answer 12

then \[\log_{4} ((x+3)*(2-x))=1\]=1

Answer 13

((x+3)*(2-x))=4^1
2x+6-x^2-3x=4
-x^2-x+6=4
-x^2-x+2=0
x^2+x-2=0
x^2+2x-x-2=0
(x+2)(x-1)=0
x=-2 or 1

Answer 14

then..\[\log_{4} x + 3 + \log_{4} 2-x = 1\]
\[\log_{4} [ (x+3)(2-x)] =1\]
which means
\[4^{1} = (x+ 3)(2-x)\]
\[4 = 6 -x -x^{2}\]
\[x^{2} + x -2 = 0\]
\[(x + 2)(x-1)=0\]
\[x = -2 \ or \ x =1\]

Answer 15

@r. gupta check again

Answer 16

got it ? :-) I think u should have no prob in understanding cause u've seem to got lot of answers bro lol

Answer 17

How do you guys cancel the \[\log_{4} \] ?

Answer 18

4^(i wrote the expression thinking its - in the question....for this question