Question

Find x in 4^x-9^x = 0

Answer 1

1-1=0 what x will give you 1

Answer 2

@zpupster :)

Answer 3

huh? I don't get it. Can you explain thoroughly. Please?

Answer 4

@zpupster

Answer 5

anything raised to the 0 is 1

Answer 6

x = 0 is the answer

Answer 7

to solve this now u need
to add 9^x to both side
4^x=9^x... now x is the same in both cases but 4 dnt equal 9
so there is no integer number (x)>0 could make this true unless u know the rule of a^0=1,,, such that a any num.

Answer 8

zpupster had the right idea except he made a human error ( x = 1 was incorrect)

Answer 9

so do you mean?

4^x - 9^x = 0
1-1=0
x=0
checking
4^0-9^0
1-1=0
0=0

Answer 10

right

Answer 11

is my solution correct?

Answer 12

yes

Answer 13

u r correct

Answer 14

@cwrw238 zpupster give 0 as an answer too !!

Answer 15

x^0 , 4^0 - anything to power 0 = 1

Answer 16

hmm. Now i see :) hehe. Thank yoou so much for your help guys. I don't know whick one of you i will award a best response

Answer 17

i have another problem. Can you help me guys? just last two. Please?

Answer 18

look.

find x if 2^(3-x) = 565

Answer 19

oops - sorry - i misread his zpupster's first post

Answer 20

find x if 2^(3-x) = 565

Answer 21

please :(

Answer 22

i really can't get the answer.

Answer 23

did you get the answer? @ikram002p

Answer 24

im trying but r u sure from 565 ??

Answer 25

ok try
take reciprocal

2^x-3 = 1/565 then
log base 2 both sides
x-3 = - log(565)/log(2)

add 3 both sides

Answer 26

yes. it's 565. :3 my geo prof. Told me to use logarithmic, coz 565 is impossible to be in exponential form of 2. But, we are not yet through with logarithmic law

Answer 27

@zpupster i don't get it. T_T

Answer 28

mmm the qs couldnt be solved without log... so u need to learn log to solve it ..