How do you solve (3/5)^x = 7^(1-x)?

Question

How do you solve (3/5)^x = 7^(1-x)?

Answer 1

(3/5)^x = 7^(1-x)?

log([3/5]^x) = log(7^[1-x])
x log(3/5) = (1-x)log(7)

x log(3/5) = log(7) - x log(7)
x log(3/5) + x log(7) = log(7)
x ( log(3/5) + log(7)) = log(7)
x ( log(21/5) ) = log(7)
x = log(7) / log(21/5)
= 1.3560

Remember... it makes NO difference if you use LOG LN or log to anyother base.... the answer WILL be the same

Answer 2

take logs...

I like natural logs......
ln [ A ^ N ] = N ln A


ln [ ( 3/5)^x ] = ln [ 7 ^ ( 1 - x) ]
x ln ( 3/5 ) = ( 1- x) ln 7
x ln (3/5) = ln 7 - x ln 7
xln(3/5 ) + x ln 7 = ln 7
x( ln (3/5) + ln 7 ) = ln 7
x = ln 7 / ( ln (3/5) + ln 7 )

or as a decimal.... x = 1.35595....

there are other ways to write this solution, if you need it in "exact " form ... ... ln( 3/5) = ln 3 - ln 5, for example can be substituted into the work above, and.. ln A + ln B = ln AB could also be used to simplify the problem... eg. ln(3/5) + ln 7 = ln ( 21 / 5), for example...etc...etc.........

Also..... regular logs... log base 10 ... will work exactly the same, and you will get:
x = log 7 / ( log (3/5) + log 7 ) = 1.35595...

hope this helps...
**********************************
don't forget to choose a Best Answer...

Answer 3

you use the logs remember log a^x = x log a

so log (3/5)^x = x* log 3/5=x*log 0.6 =-0.2218*logx
log 7^(1-x) = (1-x) log7=0.845(1-x)
so 0.845= (0.845-0.2218)*logx= 0.6233*logx
log x= 0.845/0.6233=1.356
x=22.68