Question

1. If logox - logoy = 1 and x + y = 11, then x =​

Answer 1

Answer:

Consider the given equation.

2log(x

2

y)=3+log x−log y

log (x

2

y)

2

=3+log

y

x

log (x

4

y

2

)−log

y

x

=3

log (x

4

y

2

×

x

y

)=3

log (xy)

3

=3

(xy)

3

=10

3

...........(log

a

b

=x⇒b=a

x

)

xy=10⇒y=

x

10

........ (1)

And (x−y)=3........(2)

substitute equation (1) in (2)

x−

x

10

=3

x

x

2

−10

=3

x

2

−3x−10=0

x

2

−5x+2x−10=0

x(x−5)+2(x−5)=0

(x−5)(x+2)=0

x=5 or −2

But x>0

Hence, x=5 is the answer.

Answer 2

Answer:

Given:-

${ \bullet \: \: { \sf{ log_{o}x - log_{o}y = 1}}}$

${ \bullet \: \: { \sf{x + y = 11}}}$

Find:-

Value of x??

Solution:-

${ \sf{x + y = 11.....(1)}}$

${ \sf{ log_{o}x - log_{o}y = 1 }}$

From the formula ⤵

${ \boxed{ \sf{ log \: a - log \: b = log \: \: \frac{a}{b} }}}$

${ \sf{ log_{o} (\frac{x}{y} ) = 1 }}$

We know that log 10 = 1...

${ \sf{ log_{o} (\frac{x}{y} ) = log_{o}(10)}}$

${ \to{ \sf{ \frac{x}{y } = 10}}}$

${ \to{ \sf{x = 10y}}}$

So, Now substitute the value of x in equation (1)

${ \to{ \sf{x + y = 11}}}$

${ \to{ \sf{10y + y = 11}}}$

${ \to{ \sf{11y = 11}}}$

${ \to{ \sf{y = \frac{11}{11} = 1 }}}$

So, the value of y is 1

Substitute value of y in equation (1)

${ \to{ \sf{x + y = 11}}}$

${ \to{ \sf{x + 1 = 11}}}$

${ \to{ \sf{x = 11 - 1}}}$

${ \sf{ \to{x = 10}}}$

Therefore, the value of x is 10.