Question

Given log3 = 0.4771 and log2 = 0.3010 the value of x satisfying the equation 5^x . 27^(1-x) = 0.1 where x\[>\] 0 , is?

Answer 1

the answer is 3.32 plz solve for it.....

Answer 2

is the problem this:\[5^x*27^{1-x}=0.1\]

Answer 3

yes

Answer 4

ok, first write 27 as \(3^3\) to get:\[5^x*(3^3)^{1-x}=0.1\]\[5^x*3^{3(1-x)}=0.1\]

Answer 5

ok

Answer 6

then write 5 as 10/2 to get:\[(\frac{10}{2})^x*3^{3(1-x)}=0.1\]

Answer 7

ok

Answer 8

then take logs of both sides and you should be able to solve from there

Answer 9

let me try and will inform u if i have doubts thazz

Answer 10

log to base 10

Answer 11

not getting @asnaseer

Answer 12

ok, let me describe it step by step...

Answer 13

we ended up with:\[(\frac{10}{2})^x*3^{3(1-x)}=0.1\]therefore:\[\frac{10^x}{2^x}*3^{3(1-x)}=0.1=10^{-1}\]multiply both sides by \(2^x\) to get:\[10^x*3^{3(1-x)}=10^{-1}*2^x\]

Answer 14

ok

Answer 15

now lets take logs of both sides:\[\log(10^x*3^{3(1-x)})=\log(10^{-1}*2^x)\]therefore:\[\log(10^x)+\log(3^{3(1-x)})=\log(10^{-1})+\log(2^x)\]

Answer 16

which leads to:\[x+3(1-x)\log(3)=-1+x\log(2)\]you should be able to solve from here...

Answer 17

it is 2(1-x)...@asnaseer

Answer 18

?

Answer 19

is what 2(1-x)?

Answer 20

x+3(1−x)log(3)=−1+xlog(2)

Answer 21

\[x+3(1-x)\log(3)=-1+x\log(2)\]therefore:\[x+3\log(3)-3x\log(3)=-1+x\log(2)\]\[1+3\log(3)=x\log(2)+3x\log(3)-x=x(\log(2)+3\log(3)-1)\]therefore:\[x=\frac{1+3\log(3)}{\log(2)+3\log(3)-1}\]

Answer 22

plug in values for \(\log(2)\) and \(\log(3)\) to find x.

Answer 23

ok thanzzz

Answer 24

yw