Logarithm questions

I'm stuck on the following questions - I get so far and then I don't know where to go....

1) Find x in terms of a where a > 0 and a =/= 100The logarithms are to base 10.

2 log(2x) = 1 + log a

2)

Express log(2√10) - ⅓ log0.8 - log(10/3) in the form c + log d where c and d are rational numbers and the logarithms are to the base 10.

NB: (2√10) = 2 root10

3) Find y in terms of x:

1n(x^4) - 1n(x^2y) + 1n(y^2) = 0, expressing your answer in the form y=kx^n and giving the values of the constants k and n.

Reply 1

friendsfanatic

OP

4

Oh and thanks in advance :biggrin:

Reply 2

bob_54321

i don't quite see what the first question wants.
as for 2)

log(2√10) - ⅓ log0.8 - log(10/3)
= log(2√10) - log (2/cb√10) - log(10/3)
= log(10^(5/6)) - log10/3
= log (10^(-1/6) * 3)
= -1/6 + log3

Reply 3

friendsfanatic

OP

4

Sorry I kinda forgot to explain the first Q :colondollar:

I don't get 3) at all

Reply 4

friendsfanatic

OP

4

bob_54321

i don't quite see what the first question wants.
as for 2)

log(2√10) - ⅓ log0.8 - log(10/3)
= log(2√10) - log (2/cb√10) - log(10/3)
= log(10^(5/6)) - log10/3
= log (10^(-1/6) * 3)
= -1/6 + log3

What does cb stand for?

Reply 5

Knogle

16

I tried working on #3 (and so did 2 of my friends) but none of us could reach an answer. :/ I'm going to blame that on the fact that I haven't touched 'A' level maths for well over a year, heh.

Reply 6

username9816

14

friendsfanatic

3) Find y in terms of x:

1n(x^4) - 1n(x^2y) + 1n(y^2) = 0, expressing your answer in the form y=kx^n and giving the values of the constants k and n.

ln(x^4) - ln(x^2y) + ln(y^2) = 0
=> ln(x^4y^2 / x^2y) = 0
=> ln(x^2y) = 0
=> x^2y = 1
=> y = x^(-2)

:. k = 1, n = -2

Reply 7

Knogle

16

Nima

ln(x^4) - ln(x^2y) + ln(y^2) = 0
=> ln(x^4y^2 / x^2y) = 0
=> ln(x^2y) = 0
=> x^2y = 1
=> y = x^(-2)

:. k = 1, n = -2

How did you get from ln(x^4y^2 / x^2y) = 0 to ln(x^2y) = 0?

Reply 8

username9816

14

friendsfanatic

1) Find x in terms of a where a > 0 and a =/= 100. The logarithms are to base 10.

2log(2x) = 1 + log a

2log(2x) = 1 + log a
=> log[Sqrt(2x)] - loga = 1
=> log{Sqrt(2x)/a} = 1
=> Sqrt(2x)/a = 10
=> Sqrt(2x) = 10a
=> 2x = 100a^2
=> x = 50a^2

Reply 9

username9816

14

Knogle

How did you get from ln(x^4y^2 / x^2y) = 0 to ln(x^2y) = 0?

By simplifying the fraction. Cancel x^2 on the top and bottom, and cancel y on the top and bottom

Reply 10

Knogle

16

Nima

By simplifying the fraction. Cancel x^2 on the top and bottom, and cancel y on the top and bottom

Hmmmm, one of us is seeing things wrongly.

How can you cancel x^2 at the bottom? It's x^(2y).

Reply 11

bob_54321

friendsfanatic

What does cb stand for?

i made cb up really, used it to stand for the cube root, lol. sorry.

Reply 12

friendsfanatic

OP

4

Nima

2log(2x) = 1 + log a
=> log[Sqrt(2x)] - loga = 1
=> log{Sqrt(2x)/a} = 1
=> Sqrt(2x)/a = 10
=> Sqrt(2x) = 10a
=> 2x = 100a^2
=> x = 50a^2

if it's 2log(2x) surely it'd be log(2x)^2?

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