Question

Log4 X-log16 (x+3)=1/2
The 4 and 16 are bases can someone please tell me how to solve this?

Answer 1

change of base perhaps??

Answer 2

4^y = x

16^y-3 = x would that work out ?

Answer 3

change both bases to 4, then use the laws of exponents to combine\[\log_{16}(x+3)=y\] means
\[16^y=(x+3)\]
\[4^{2y}=x+3\]
\[2y=log_4(x+3)\]
\[y=\frac{log_4(x+3)}{2}\]

Answer 4

4^y - 4^(2y-6) = 1/2

Answer 5

i was close lol

Answer 6

change of base gets me two joined equations with a factored out x

Answer 7

log4(x) - [log4(x+3)/log4(16)] = 1/2

log4(x) - (1/2)log4(x+3) = 1/2

Answer 8

idk if i can change both bases to 4

Answer 9

log4(x/sqrt(x+3)) = 1/2

Answer 10

you can do that?

Answer 11

sqrt(4) = sqrt(x/(x+3)) i believe so :)

Answer 12

then exponentiate to get \[\frac{x}{\sqrt{x+3}}=4^{\frac{1}{2}}\]

Answer 13

4 = x/x+3

4x+12 = x

3x = -12

x = -4...maybe

Answer 14

i got lost in it for a moment....

Answer 15

no way -4 is an answer since you cannot take log of -4

Answer 16

i know i know.... my brain took a momentary leave of absence lol

Answer 17

the 4 is not negative

Answer 18

sqrt(4) = x/(x+3) is what i meant to type :)

Answer 19

4 = x^2/x^2+9+6x

Answer 20

4x^2 +24x +36 = x^2

3x^2 +24x +36 = 0... thats better me thinks

Answer 21

3(x^2 +8x +12) = 0

Answer 22

x = -6 and -2 by that accounting...... so maybe not ;)

Answer 23

oops
start with \[\frac{x}{\sqrt{x+3}}=4^{\frac{1}{2}}\]

Answer 24

...... yeah, it would help if I did the actual problem instead of making up me own lol

Answer 25

\[x=2\sqrt{x+3}\]

Answer 26

\[x^2=4(x+3)=4x+12\]
\[x^2-4x-12=0\]
\[(x-6)(x+2)=0\]
\[x=6, x=-2\]
-2 not a solution. we can try 6 to see if it works.

Answer 27

yup 6 it is

Answer 28

hey where did the log16 go?

Answer 29

we..and be that i me sat.... modified it to a log4(x+3)/2

Answer 30

I get it its just i do not know where the log16 went
the log of16 is 1.20411

Answer 31

change of base just changes the appearance and not the value

Answer 32

log4(16) not log(16)

Answer 33

let us go slow. \[log_{16}(x)=y\] means \[16^y=x\]
for example
\[log_{16}(4096)=3\] because \[16^3=4096\]

Answer 34

but \[16=4^2\] so it is also true that \[(4^2)^3=4069\]
\[4^6=4069\]
so \[log_4(4069)=6\]

Answer 35

so the log base 16 did not disappear, just converted to log base 4

Answer 36

oho i get it thank you
:]

Answer 37

i got X=2.6667

Answer 38

how did you get that? i am fairly certain x = 6