the smaller of two numbers is 2 less than 1/3 the larger. the sum of the two numberrs is 50 what is the value of the numbers

Mathematics

1 answer:

Alja [10]2 years ago

8 0

The smaller number is 14 and the larger number is 36.

<h3>What is the equation?</h3>

The equation is defined as mathematical statements that have a minimum of two terms containing variables or numbers that are equal.

Let the smaller number would be x

And the larger number would be y

Given that the smaller of two numbers is 2 less than 1/3 the larger

So x - 2 = (1/3)y ⇒ 3x - 6 = y

The sum of the two numbers is 50

So x + y = 50

Substitute the value of equation 3x - 6 = y in the above equation,

⇒ x + 3x - 6 = 50

⇒ 4x = 50 + 6

⇒ x = 56/4

⇒ x = 14

Substitute the value of equation x = 14 in the equation x + y = 50,

⇒ 14 + y = 50

⇒ y = 36

Hence, the smaller number is 14 and the larger number is 36.

Learn more about the equation here:

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Express this to single logarithm

zzz [600]

Answer: $\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right)$

We have something in the form log(x/y) where x = q^2*sqrt(m) and y = n^3. The log is base 2.

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Explanation:

It seems strange how the first two logs you wrote are base 2, but the third one is not. I'll assume that you meant to say it's also base 2. Because base 2 is fundamental to computing, logs of this nature are often referred to as binary logarithms.

I'm going to use these three log rules, which apply to any base.

log(A) + log(B) = log(A*B)
log(A) - log(B) = log(A/B)
B*log(A) = log(A^B)

From there, we can then say the following:

$\frac{1}{2}\log_{2}\left(m\right)-3\log_{2}\left(n\right)+2\log_{2}\left(q\right)\\\\\log_{2}\left(m^{1/2}\right)-\log_{2}\left(n^3\right)+\log_{2}\left(q^2\right) \ \text{ .... use log rule 3}\\\\\log_{2}\left(\sqrt{m}\right)+\log_{2}\left(q^2\right)-\log_{2}\left(n^3\right)\\\\\log_{2}\left(\sqrt{m}*q^2\right)-\log_{2}\left(n^3\right) \ \text{ .... use log rule 1}\\\\\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right) \ \text{ .... use log rule 2}$