simultaneous logarithmic??

Question

hi, im just wonder how do i go about answering this question?

log x + log y = log 3
log (3x +y) = 1

thanks bhavinp

Answer 1

We have:
log(xy) = log 3
So, xy = 3

log (3x + y) = log 10
So, 3x + y = 10

Substituting y = 3/x into the 2nd equation, rearranging & factorising:
3x + 3/x = 10
3x^2 - 10x + 3 = 0
3x^2 - 9x - x + 3 = 0
3x(x - 3) - (x - 3) = 0
(3x - 1)(x - 3) = 0

Giving solutions for x:
x = 1/3 and x = 3

Substituting the value of x into xy = 3:
When x = 1/3, y = 9
When x = 3, y = 1

Answer 2

logx + log y = log(xy) = log 3

therefore, xy = 3.

log( 3x + y) = 1

the log of the base itself is 1( that is log a to base a is 1).

therefore, 3x + y = 10

xy = 3

3x + y = 10

(3x + y)^2 = 9x^2 + 6xy + y^2


substituting xy = 3,

9x^2 + 18 + y^2 = (3x + y)^2

9x^2 + 18 + y^2 = 10^2

9x^2 + 18 + y^2 = 100

9x^2 + y^2 = 100 - 18
= 82

we use this to find (3x - y)^2 since we know both xy and 9x^2 + y^2.

substiututing these values we find that 3x - y = plus or minus 8.

we have 3x + y = 10 and 3x - y = plus or minus 8.

solve this and you get x and y.

Answer 3

log x+log y =log xy =log 3 so xy=3

log(3x+1) =1 this means 3x+1 = 10

so 3x=9
x=3 and y=1

proof log3+log1 = log3 yes since log 1=0
log (3*3+1) = log 10 =1

Answer 4

log x + log y = log 3
=> log xy = 3
=> xy = 3 .............................(i)

log (3x +y) = 1
=> 3x + y = 10 (anti log 1= 10 or log 10 = 1) assuming a base of 10
=> y= 10- 3x......................... (ii)

by (i) & (ii)
x ( 10 - 3x) = 3
=>10 x - 3x^2 = 3
=> - 3 x^2 + 10x - 3 = 0
=>3 x^2 -10 x + 3 = 0
=>3 x^2 -9 x - x + 3 = 0
=>3x(x-3) –1(x-3) = 0
=>(3x-1) (x-3) = 0
=>x= 1/3, x= 3……………. (iii)

by (i) xy = 3
=> when x= 1/3, y = 9
& when x = 3, y = 1

Answer 5

by the rules of logs, you can reduce log x + log y to log (xy), so xy =3
then you can also say x = 3/y so substitute that into the bottom equation and you get
log ( 3 (3/y) + y) = 1
assuming your logs are to the base 10 as is standard, you can make them exponential powers to get rid of the logs
so 10^log(3(3/y) + y) = 3(3/y)+y
and 10^1 is 10
so 10=3(3/y) + y
this can be changed into a simple quadratic equation by multiplying everything by y and then moving it all to one side.
y^2 -10y +9 = 0
so y is either 9 or 1,
which give you the options for x (remember xy=3?) to be 1/3 or 3 respectively.

Answer 6

You need to state which base your logs are in..


BUT since you have written log rather than ln ..I shall assume base10

All you need to remember is this
if log m=n
then m=10^n

AND

log a + log b= log ab

so eq1
log x +log y= log xy=log 3

so xy=3

eq2
log 3x+y= 1

so 3x+y= 10


Now go solve!

Answer 7

log x+log y :: log 3
log xy :: log 3
xy :: 3