Question

WHERE IS MY MISTAKE? :SS

4^x - 2^x - 12 = 0
4^x - 2^x = 12
log(4^x) - log(2^x) = log(12)
x*log(4) - x*log(2) = log(12)
x*(log(4) - log(2)) = log(12)
x*log(2) = log(12)
x = log(12) / log(2)
x = log_2(12)
2^x = 12


WHERE IS MY MISTAKE TILL NOW?

Answer 1

line one

Answer 2

what is it

Answer 3

you cannot take the log of both sides by taking the log separately

Answer 4

in other words if \(x+y=z\) it is not true that \[\log(x)+\log(y)=\log(z)\]

Answer 5

what is it was x + y - z = 0 ?

Answer 6

what if*

Answer 7

Convert the 4^x and 2^x to be expressions of the same base...

Answer 8

the gimmick here is to rewrite
\[4^x+2^x-12=0\] as
\[2^{2x}-x^x-12=0\] and solve the quadratic equation

Answer 9

what if it was x + y - z = 0 ?

Answer 10

for example if you call \(y=2^x\) then you have
\[y^2-y-12=0\]

Answer 11

you cannot take the log piece by piece
here is a simple example
we know \(3+7=10\)
taking the log we have
\[\log(3+7)=\log(10)\] not
\[\log(3)+\log(7)=\log(10)\]

Answer 12

that applies for nl(x) too?

Answer 13

yes

Answer 14

\[\log(a+b)\neq \log(a)+\log(b)\]

Answer 15

ok please solve me this 2e^-x + 1 = 5

Answer 16

I need to see something

Answer 17

true of almost any function right?
\[\sqrt{9+16}\neq \sqrt{9}+\sqrt{16}\]

Answer 18

yes can you please solve me this, its an easy one, I just need to meisure a small logic that's behind

Answer 19

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

Answer 20

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

Answer 21

2e^(-x) + 1 = 5 sorry

Answer 22

your first goal is to make it look like
\[e^{-x}=\text{something}\] so you can solve by taking the log

Answer 23

first step is to subtract 1 to get
\[2e^{-x}=4\]

Answer 24

divide by 2 and get
\[e^{-x}=2\]

Answer 25

in equivalent logarithmic for that means
\[-x=\ln(2)\] and so
\[x=-\ln(2)\]

Answer 26

Then it goes x = ln(2^-1)

and x = ln(1/2) :P

Answer 27

if you like, yes
they are the same number

Answer 28

\[\log(\frac{1}{a})=-\log(a)\]

Answer 29

however your first problem does not really involve logs

Answer 30

log(a^-1)

Answer 31

\[y^2-y-12=0\]
\[(y-4)(y+3)=0\]
\[y=4, y=-3\]
\[2^x=4\] or \[2^x=-3\] the second one is not possible, and if
\[2^x=4\] then \[x=2\]

Answer 32

please give me the first step to solve (e^x - e^-x)/2 = 2