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Answer 1

@eashmore der ?? help plzz

Answer 2

\[\Large x \log_{2}2 + \log_{2}2 - \log_{2}(2^{x} + 3)^{2} + \log_{4}(10 - 2^{x})^{2} = 0\]

how do you turn log_4 to log_2 again?

Answer 3

see I have been stupidly solving this actually, when you can just use trial and error with each option, i guess that will be better, shorter, and wittier.

Answer 4

\[\large \log_{2}2^{x} + \log_{2}2 - \log_{2}(2x+3)^{2} + \frac {\log_{2}(10 - 2^{x})^{2}}{\log_{2}4} = 0\]

anyone has any idea where to from here??

Answer 5

log laws since the base is 2..

Answer 6

combine them..

Answer 7

Or you can convert them into a simplified form before initiating the trial and error.

Answer 8

but you can't always do trial and error if it is not a mulptiple choice..

Answer 9

I don't even.....

Answer 10

\[\large log_{2}\frac {(2^{x})(2)}{(2x+3)^{2}} + \frac {log_{2}(10 - 2^{x})^{2}}{log_{2}4} = 0\]

right?

Answer 11

I 'll have to restart my PC for the darn scannner won't work, so please excuse me for a few minutes :p

Answer 12

ha! i got it >:)))

Answer 13

Actualy I don't know how you got the first step..

Answer 14

I'm very bad at logs
can you write down the laws ?

Answer 15

*which you used here?

Answer 16

\[\large log_{2}\frac{(2^{x})(2)}{(2x+3)^{2}} + \frac {log_{2}(10 - 2^{x})^{2}}{log_{2}2^{2}} = 0\]

\[\large log_{2}\frac{(2^{x})(2)}{(2x+3)^{2}} + \frac {log_{2}(10 - 2^{x})^{2}}{2} = 0\]

\[\large log_{2}\frac{(2^{x})(2)}{(2x+3)^{2}} + \frac{1}{2}log_{2}(10 - 2^{x})^{2} = 0\]

\[\large log_{2}\frac{(2^{x})(2)}{(2x+3)^{2}} + log_{2}(10 - 2^{x}) = 0\]

\[\large log_{2}\frac{(2^{x})(2)(10 - 2^{x})}{(2x+3)^{2}} = 0\]

is it becoming obvious now? :DDD

Answer 17

I think that you have to check your first step..

Answer 18

me mimi? what did i do wrong?

Answer 19

ooohh 2^-x? sorry....but it doesnt matter so far since i havent touched it yet...

Answer 20

Okay "here we are.."

Answer 21

But, I am wrong somewhere, since none of the options seem to suit it. :( whyamisocareless

Answer 22

Lol let me see

Answer 23

\[\large 2^{0} = \frac {(2^{x})(2)(10 - 2^{-x})}{(2x + 3)^{2}}\]

\[\large 1 = \frac {(2^{x})(2)(10 - 2^{-x})}{(2x + 3)^{2}}\]

this just got hard :/

Answer 24

Lgba, you going fine.

Answer 25

let's just go with apoor's answer :DDD

Answer 26

Wait lgb i seriously didn't understand your first step ,i'm not good in log (specially with base)

Answer 27

Ya, flash those teeth, when you know I'm not getting an answer. :/

lol

Answer 28

which one of them diya?
xlog_2(2) i got because log_2 (2) = 1 then 1(x) = x so xlog_2(2) =(1)(x) = x got it?

the next one is log_2 (2) = 1 (logarithmic property)

the next ones are power rules where x log_a (b) = log_a (b)^x

make sense?

Answer 29

Diya, \[\huge\log _{a^n}^{x} = (1/n)\log _{x}^{a}\]

Answer 30

that's all i did in the first step :D

Answer 31

\[\huge\log _{a}^{x^n} = n*\log _{a}^{x}\]

Answer 32

lol my variables screwed up//

Answer 33

\[\huge logA ^{n} = n*logA\]

Answer 34

I get confused when there is base

Answer 35

\[\Huge log_{a}(x)^{n} = n log_{a}(x) \]

better...that's what you meant @apoorvk :P

Answer 36

okay...let's list all the theorems/properties involved here...wait...

Answer 37

same thing with or without the base..

Answer 38

Yeah, that's what I meant. Definitely better @lgbasallote :)

Answer 39

\[\Huge log_{a}a = 1\]
\[\Huge log_{a}b + log_{a}c = log_{a}(bc) \]
\[\Huge log_{a}b - log_{a}c = log_{a}\frac {b}{c}\]
\[\Huge log_{a}b = x\] is equal to \[\Huge a^{x} = b\]

Answer 40

the power rule was stated above already

Answer 41

Cool!

Answer 42

Diya diya diya, wait.
try to think what is happening in here.

suppose 2^x gives 64. you know x=6

now now now, what if I replace that 2 with 4, which is 2^3? then x=2 (which is 6/2). So do you notice that 'x' gets divided by the power which I am raising '2' to.

Answer 43

sooo...im still stuck with MY last step -__-

Answer 44

that was what happened in the the first step. we 'played' with the base. We do that every time, so as to make the base of the log terms same. So that we can apply use properties, and finish off the problem!

Answer 45

let me go through it

Answer 46

oh yeah..i used another property...

\[\Huge \log_{a}b = \frac {log_{c}b}{log_{c}a}\]

Answer 47

CORRECTION>>>>>> I am truly at the epitome of carelessness.

Diya diya diya, wait.
try to think what is happening in here.

suppose 2^x gives 64. you know x=6

now now now, what if I replace that 2 with 4, which is 2^3? then x=2 (which is 6/'3'). So do you notice that 'x' gets divided by the power which I am raising '2' to.

Answer 48

\[\Huge \mathbf{L} \mathsf{A} \mathbf{T} \mathsf{E} \mathbf{X}\]

lol sorry for that =)))) my mind is so dizzy

Answer 49

CORRREECTIONN ALERT!!!!! @Diyadiya @lgbasallote @Mimi_x3

Mr. Carelessness, aka @apoorvk made a small tiny-winy mistake there, as always.
In that scan, at the end of the simplification process, there is a term circled in red. In there, the denominator should read:
\[(t+3)^2\]

AND NOT:
\[(t+3)^3\]

So that should solve the problem.


Or does it? Still none of the options satisfy 't'. In fact if you notice carefully, 't' cannot be a whole no. at all. Can you say why? ;-]

Numerically solving the equation, got some very irrational values for 't'

\[t = 2^x = log_2 (7 + \sqrt6) , log_2 (7 - \sqrt6) \]


So basically, there was some silly typing error in the question I guess. :/

Answer 50

Haha, I tried tagging myself, and it worked =D