Question

A logarithmic equation. I got it however , but i wish to share it with you all @mathslover :)
Anyone could try to solve this problem

Answer 1

\[\huge x ^{1+logx} = 10x\]

Answer 2

@mathslover

Answer 3

Anyone may solve this if they want . I got it

Answer 4

I posted this this because some of them are doing logarithms and i am doing it too

Answer 5

Its a tricky one. somewhat

Answer 6

i thought this was for mathsolver :P

Answer 7

Anyone here i specially tagged him because he is doing log

Answer 8

ok ill give a direct answer :P

Answer 9

yes

Answer 10

message me

Answer 11

It is not so easy as it looks

Answer 12

Its easy...

Answer 13

@No.name - I think, the first step will be to take log both sides.

Answer 14

yes

Answer 15

Is the answer 0 ?

Answer 16

zero dnt count :)

Answer 17

O.O yeah.. I meant to say 1

Answer 18

Sorry, i calculated log x = 0

So, x = 1

Answer 19

haha no lol
if x=1
then
1=10 O.o

Answer 20

o.O leave it

Answer 21

hmm, lol, is it 10

Answer 22

i have solution but its irrational xD

Answer 23

Hmm... I am getting 10 @No.name

Answer 24

cool it work :O

Answer 25

(1 + log x) (logx ) = log (10x)

log x + (log x)^2 = log x + 1

(log x)^2 = 1

log x = +/- 1

log x can not be -1...

So, log x = 1

thus, x = 10

Answer 26

why its work mmm

Answer 27

there are 2 answers

Answer 28

One of them is 10?

Answer 29

that's why its tricky

Answer 30

yes one is 10

Answer 31

hmm ... we might want to take the conditions here?

Answer 32

nope its very straightforward

Answer 33

Oh...

Answer 34

\(\large x^{\log_{10} 10 + \log_{10} x} = 10x\)

\(\large x^{\log_{10} 10x} = x^{\log_{x} 10x}\)


\(\large \log_{10} 10x = \log_{x} 10x\)

\(\implies x = 10, ~0.1\)

Answer 35

@ganeshie8 got it !

Answer 36

One more question i have in log

Answer 37

o.O yeah, I MUST not have avoided this : log (x ) = -1 and log(x) = 1

log(x) = -1

when ... x = 10^{-1} = 1/10

Answer 38

OH.. :( I was almost there... but anyways, @ganeshie8 got it... great work

and nice quest. @No.name :-)

Answer 39

I went wrong at this pt. : "log x can not be -1..."

it can be -1 :(

Answer 40

yes