Question

3^(2x) -5(3^x)=-6

Answer 1

\[3^{2x} -5 (3^{x}) = -6 \]

Answer 2

x=1 & x=0.631

Answer 3

but how ? T.T

Answer 4

start with
\[3^{2x}-5\times 3^x+6=0\] then factor as
\[(3^x-3)(3^x-2)=0\] so either
\[3^x-3=0\implies 3^x=3\implies x=1\] or
\[3^x-2=0\implies 3^x=2\implies x=\frac{\ln(3)}{\ln(2)}\]

Answer 5

yayyyy :) I am waiting for u

Answer 6

ohh

Answer 7

log 3/ log 2 ?

Answer 8

gimmick is to think of this as a quadratic equation in
\(3^x\) if you like you can replace \(3^x\) by \(z\) and start with
\[z^2-5z=-6\]
\[z^2-5z+6=0\]
\[(z-3)(z-2)=0\] etc
then after solving to get \(z=3\) or \(z=2\) replace \(z\) by \(3^x\)

Answer 9

the first solution \(3^x=3\) is easy, because clearly \(x=1\) for the second solution you need to know that if \(b^x=A\) then \(x=\frac{\ln(A)}{\ln(b)}\)

Answer 10

ln is log or natural log ?

Answer 11

actually it makes no difference, you can use any log you like

Answer 12

it is sometimes knows as the "change of base formula" but in practical value your calculator has only two logs, log base e denoted by \(\ln\) and log base 10 denoted by \(\log\)

Answer 13

then how did he/she got x=1 & x=0.631 ??

Answer 14

so you could write
\[3^x=2\iff x=\frac{\log(2)}{\log(3)}\] or you could write
\[3^x=2\iff x=\frac{\ln(2)}{\ln(3)}\] you can check for your self that you get the same number in either case

Answer 15

and 1 ?

Answer 16

ohhhhhhhh

Answer 17

oh damn i made a typo in my first answer sorry. it should be
\[x=\frac{\ln(2)}{\ln(3)}\]

Answer 18

the 1 is from your eyeballs

Answer 19

\[3^x=3 \iff x = 1\]

Answer 20

thanks alottttttttttttttt <3

Answer 21

yw

Answer 22

\[\log_{5} ( \log_{3} x) =0\]

Answer 23

we are solving for \(x\)?

Answer 24

yahh

Answer 25

well not matter what the base is, we know \(\log_b(1)=0\) which means we must have \(\log_3(x)=1\) and therefore \(x=3\)

Answer 26

sorry tmr I will have a test :( so there are too many question :(

Answer 27

that one was not too bad

Answer 28

yahhhh,, I'm so crazy =))

Answer 29

good luck on your test!

Answer 30

log(35-x^3)/ log(5-x) = 3

Answer 31

thank XD, but I don't think I will do it well :$

Answer 32

\[\log(35-x^3)=3\log(5-x)\]
\[\log(35-x^3)=\log((5-x)^3)\]
\[35-x^3=(5-x)^3\]

Answer 33

and then expand it ?

Answer 34

then you have a raft of algebra to do, but don't worry, the cubes will go and you will get an easy quadratic to solve

Answer 35

yes, expand.
the right hand side will be \(-x^3+15 x^2-75 x+125\) so you have
\[35-x^3=-x^3+15 x^2-75 x+125\]

Answer 36

I did expand it, but then do I need to factor it ? or what

Answer 37

then you get
\[15x^2-75x+90=0\] and go from there

Answer 38

ohhhhhhh yahhh :) i did a mistake... that's why :(((

Answer 39

divide by 15 get
\[x^2-5x+6=0\] so it is a lot like the last one

Answer 40

gotta run, good luck

Answer 41

okayyy thanks againnnnn :)

Answer 42

\[y =\log_{2} (5x+4)\]