Question about logs?

Question

log(base9)(xy²) = 1/2 log(base3)x+log(base3)y (<----given)

solve the simultaneous equations:

log(base9)(xy²)=1/2

(log(base3)x)(log(base3)y) = -3

pls show working thanks alot

Answer 1

log(base 9)(xy²) = 1/2  

=&gt;   log(base3)(xy²) = 2 * 1/2   (change of base from 9 to 3)

=&gt;   log(base3)x + 2*log(base3)y = 1

let log(base3)x = A

let log(base3)y = B

=&gt; A + 2B = 1         ---------Eq(1)

=

(log(base3)x)(log(base3)y) = -3

=&gt; AB = -3

=&gt; B = -3/A            ---------Eq(2)

putting Eq(2) in Eq(1)

A - 6/A = 1

=&gt;  A^2 - 6 = A

=&gt; A^2 - A - 6 = 0

by factorising

(A + 2)(A - 3) = 0

there fore   A = -2; A = 3

A = -2 gives infeasible solution (hence ignore)

If A = 3 we have B = -1 from Eq(2)

thus

A  = 3   =&gt;   log(base3)x = 3   =&gt; x = 3^3 = 27    

B  = -1  =&gt;   log(base3)y = -1  =&gt; y = 3^-1 = 1/3

final solution

x = 27

y = 1/3

Note:  

the rule

(log(baseA)x)(log(baseA)y) =  log (baseA)(x * y)

is Wrong as claimed by &quot;K5&quot; in other answer...

correct

(log(baseA)x)(log(baseA)y)   =  is NOT =  log (baseA)(x * y)

Answer 2

log(base9)(xy²)=1/2 can be simplified by this rule

log (base a) (x) =y ---&gt; a^y = x

therefore,   9^(1/2) = xy² now 9^(1/2) means the sqrt of 9 which is equal to 3, or better still 9^(1/2) = (3²)^(1/2) = 3

so that means 3 =xy² which is the first equation

then

(log(base3)x)(log(base3)y) = -3

if simplified using this rule:

(log(baseA)x)(log(baseA)y) ----&gt; log (baseA)(x * y)

so equation 2 becomes:

log (base 3) (x * y) = -3

now using the first rule used in equation 1, we arrive at

3^ (-3) = xy ------&gt; Equation 2 note that 3 ^ (-3) = 1/27

so now we have the equations simplifed thus:

Equation 1---&gt; 3 = xy²

Equation 2---&gt; 1/27 =xy

so we manipulate equation 1 to get a variable of x in y thus:

x = 3/y² ----&gt; equation 4

now fix equation 4 into equation 2 to get:

1/27 = (3/y² ) (y) two y&#039;s cancel out to get 1/27 = 3/ y

now by change of subject,

y = 3 * 27 = 81

so if form equ 4,  x = 3 /y² then from y =81, x = 3 / (81)²  

which gives x = 1/2187

so y = 81, x = 1/2187 yeah

sorry i made a mistake and this makes my whole answer bougs

the rule rather is

Log (base A) x + log (base A) y = log (base A) (x *y)