another logarithm problem!?

Question

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

(that is .. 3 divided by 5 to the x = 7 to the 1-x)


.. this is what ive done so far, im not sure if its correct, but im stuck! please help! ..

ln (3/5)^x = ln 7^1-x
x ln (3/5) = 1-x ln 7
x ln 0.6 = 1-x ln 7


.. the answer from the book is .. ln 7/ln 0.6 + ln 7

Answer 1

It looks right - be more confident about your skills.

Leaving off from your last step:
x*ln(0.6) = (1-x)*ln(7)
x*ln(0.6) = ln(7) - x*ln(7)
x(ln(0.6) + ln(7)) = ln(7)
x = ln(7)/(ln(0.6) + ln(7))

That is the answer in your book.

Answer 2

xln(3/5) =(1-x)ln7
so
x(ln(3/5)+ln 7)=ln7
so x= (ln7 /(ln0.6+ln7) as 3/5 =0.6

Answer 3

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

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

x ln3/5 = (1-x) ln7

distibute
x ln(3/5) = ln7^1 - xln7

x ln(3/5) = ln7 - xln7

add xln7 for both sides
x ln(3/5) + xln7 = ln7

take out x
x (ln3/5 + ln7) = ln7
x (ln4.2) = ln7

x = ln7 / ln4.2

Answer 4

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

ln(3/5)^x = ln 7^(1-x)
x ln(3/5) = (1-x) ln(7)
x ln(3/5) = ln(7) - xln(7)
xln(3/5) + xln(7) = ln(7)
x(ln.6) + ln(7) = ln(7)
x = ln(7) / (ln(.6) + ln(7))

Answer 5

Brackets are your friends!

x ln(3/5) = (1-x) ln 7
→ x ln0.6 = ln7 - x ln7
→ x(ln0.6 + ln7) = ln7
→ x = ln7 / (ln0.6 + ln7)

Answer 6

Another logarithm problem!?
(3/5)^x = 7^1-x


(3/5)^x = 7^1-x
take Logs of both sides : xlog (0.6)=(1-x)log 7
or x[log(0.6)+log7] =log7or x=log7/[log0.6 +log 7
or x[log(4.2)] =0.845
or x=0.845/0.623=1.356 answer






90