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3 years ago

If x=log3/5,y=log5/7,z=log root7/3,find the value of x+y+z

Mathematics

1 answer:

maxonik [38]3 years ago

4 0

Answer:

Given:

x=log(3/5)

y=log(5/7)

z=log(sqrt(7/3)

Find x+y+z

Steps:

Using the law of logarithms, log A + log B = log (AB), we get

x+y+z

=log(3/5)+log(5/7)+log(sqrt(7/3)

=log(3/5*5/7*sqrt(7/3))

=log(3/7*sqrt(7/3)

=log((sqrt(3/7)*sqrt(3/7)*sqrt(7/3))

=log((sqrt(3/7)*sqrt(3/7*7/3))

=log(sqrt(3/7)*1)

=log(sqrt(3/7))

Now apply another law of logarithms, log(sqrt(A)) = (1/2) log(A)

=(1/2)log(3/7)

which is the final answer.

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Passing through (2, - 2) and perpendicular to the line whose equation is y= 5x+2

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Step-by-step explanation:

Hey there!!!

Here,

Given, A line passes through point (2,-2) and is perpendicular to the y= 5x+2.

The equation of a straight line passing through point is,

$(y - y1) = m1(x - x1)$

Now, put all values.

$(y + 2) = m1(x - 2)$

It is the 1st equation.

Another equation is;

y = 5x +2........(2nd equation).

Now, Comparing it with y = mx + c, we get;

m2=5

As per the condition of perpendicular lines,

m1×m2= -1

m1 × 5 = -1

Therefore, m2= -1/5.

Keeping the value of m1 in 1st equation.

$(y + 2) = \frac{ - 1}{5} (x - 2)$

Simplify them.

$5(y + 2) = - x + 2$

$5y + 10 = - x + 2$

$x + 5y + 8 = 0$

Therefore the required equation is x+5y+8= 0.

Hope it helps...

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