Question

Find the product of all values of "x" that satisfy the equation:-

Answer 1

\[\huge \bf x^{x^2-25}=1\]

Answer 2

@ganeshie8

Answer 3

@UnkleRhaukus

Answer 4

5 and -5

Answer 5

My my, we have a problem :D

Also, @Abmon98 we're kind of asked for the product of all the solutions to the equation, so...

Answer 6

So the answer is... 25

HAHA

^.^

Answer 7

why its 25,not -25

Answer 8

Because I said so, and I'm awesome ^.^

Answer 9

please tell me the reason

Answer 10

Silly mayankdevnami, did you notice that 1 and -1 are solutions too? ^.^

Answer 11

take log both sides may be..

Answer 12

yeah! you are right

Answer 13

\[\Huge f(x) ^{g(x)}=1\]
It'd work with
g(x) = 0 and f(x) not zero

OR

f(x) = 1 and g(x) whatever.

Answer 14

thnx... @PeterPan

Answer 15

aw yea :D

Answer 16

brilliant !

Answer 17

and what about your method. @ganeshie8

Answer 18

forget it, its a dumb method

Answer 19

please tell

Answer 20

ive told already,
take log both sides..

Answer 21

(x^2-25) log x = 0

x = +- 5
x = +-1

Answer 22

anyways, log cant digest negative values as input... so u need to be bit careful... before concluding..

Answer 23

correct man

Answer 24

thnx....

Answer 25

just noticed that \[\Huge f(x) ^{g(x)}=1\] is easily verifiable as per @ganeshie8
:3

\[\huge g(x) \log f(x) =0\]\[\huge\implies g(x) = 0 \qquad or\qquad f(x) = 1\]

Answer 26

ahh yes.. but it is not giving us -1 as solution.... idk how it got disappeared :o

Answer 27

Okay, let's change it a bit :)
\[\Huge f(x) ^{g(x)}=1\]
means:
f(x) not zero and g(x) = 0
OR
f(x) = 1 and g(x) whatever
OR
f(x) = -1 and g(x) = 0(mod 2)

better?

Answer 28

ur method is okay

im talking about taking log..

Answer 29

when we take log, x = -1 solution is getting dropped off..

Answer 30

(x^2-25) log x = 0

x = +- 5
x = + 1

Answer 31

this gives oly 3 solutions,
but clearly -1 is also a solution... but putting -1 inside log is illegal uhmm

Answer 32

@ganeshie8
I got it :D

Answer 33

tell me lol ive been trying and almost giving up xD

Answer 34

\(x^2 - 25\) is either an even integer or it is not an even integer.
Suppose it is not an even integer.
Then
\[\Huge x ^{x^2-25}=1\] only has x=1 for its solution.

Suppose it IS an even integer.
Then...
\[\Huge x ^{x^2-25}=\color{blue}{|x|^{x^2-25}}=1\]
implying
\[\huge (x^2-25)\log|x|=0\]
and work from there, this DOES give x=-1 as a solution

^.^

Answer 35

beautiful !!!!! makes perfect sense now ty lol xD

Answer 36

;)

Answer 37

No, wait, change again...My previous attempt is awkward.

Start from here.
\[\Large (x^2-25)\log(x)=0\]
Multiply 2 to both sides.
\[\Large \color{green}2(x^2-25)\log(x)=\color{green}2\times 0\]\[\Large(x^2-25)\log(x^2)=0\]
\[\Large \implies x^2-25 = 0 \qquad or \qquad x^2 = 1\]

Much eleganter ^.^

Answer 38

Hey @ganeshie8 that thing up there looks way better than my previous two-case method, which was a bit iffy, now that I think about it XD

Answer 39

yupp this looks elegant and simpler :D

Answer 40

\[\begin{align}
x^{x^2-25}&=1\\
\log_x|x^{x^2-25}|&=\log_x|1|\\
(x^2-25)\log_x|x|&=0\\
&&x^2-25&=0&\text{or}&&\log_x|x|&=0\\
&&x^2&=25&&&x^{\log_x|x|}&=x^0\\
&&x&=\pm\sqrt{25}&&&|x|&=1\\
&&x&=\pm5&&&x&=\pm1
\end{align}\]
\[x_{1,2,3,4}=\pm1,\pm5\]
\[x_1x_2x_3x_4=(1)(-1)(5)(-5)=25\]