Question

0.5^x-1 = 0.4^2x

Answer 1

0.5^x-1=0.4^2x
0.5^x-0.4^2x=1
o.5^x-0.16^x=1

Answer 2

do you use log?

Answer 3

@sauravshakya

Answer 4

I am thinking

Answer 5

@ganeshie8 Can u solve this

Answer 6

hey i think the question has -1 on exponent

\(0.5^{x-1} = 0.4^{2x}\)

then we can use log ?

Answer 7

what do u thin @sauravshakya

Answer 8

may i know answer please?

Answer 9

Oh ya

Answer 10

Ya @ganeshie8 ....... I think the question is the that

Answer 11

the answer in the textbook is -0.608

Answer 12

0.5^(x-1)=0.4^2x
0.5^(x-1)=0.16^x
log0.5^(x-1)=log(0.16^x)
(x-1)log0.5=xlog0.16
xlog0.5 -log0.5 = xlog0.16
x(log0.5-log0.16)=log0.5
x=log0.5/(log0.5/log0.16)

Answer 13

Can u use calculator to find that and check...... cause I dont have a calculator

Answer 14

-0.796

Answer 15

is it division? @sauravshakya can you teach me logs?

Answer 16

If a^b=c then it can be written as

Answer 17

got it? @muhammad9t5

Answer 18

if you don't mind do you Skype?

Answer 19

would it start off like this instead?
0.5^(x-1) = 0.4^(2x)
log0.5^(x-1) = log0.4^(2x)
(x-1)log0.5 = (2x)log0.4

Answer 20

Yes u can do that

Answer 21

yes sir are you there? @sauravshakya

Answer 22

I am not good at teaching stuff...... U can have a look at KHANACADEMY

Answer 23

hhhmm do you have learned from that?

Answer 24

Yep

Answer 25

And books

Answer 26

then?
x-1 = (2x) (log0.5/log.0.4)

Answer 27

(x-1)log0.5 = (2x)log0.4
x log0.5 -log0.5=2xlog0.4
xlog0.5 - 2xlog0.4 = log 0.5