Video Transcript
Find the value of the constant 𝑘,
given that four 𝑥 raised to the fifth power 𝑦 squared plus 𝑘𝑥 raised to the
seventh power 𝑦 cubed all over two 𝑥 squared 𝑦 is equal to two 𝑥 cubed 𝑦 plus
three 𝑥 raised to the fifth power 𝑦 squared.
In this question, we are asked to
find the value of a constant 𝑘 given an equation involving this unknown
constant. To answer this question, it is
worth recalling that when we say “equals” in this case, we are really asking for the
expressions to be equivalent. In other words, we want both sides
of the equation to be equal for any valid values of 𝑥 and 𝑦.
We can find the value of 𝑘 by
evaluating the division on the left-hand side of the equation. To do this, we will start by
dividing each term in the numerator separately by the denominator. We can then simplify by recalling
that the quotient rule for exponents tells us if a base 𝑏 is nonzero, then 𝑏
raised to the power of 𝑚 over 𝑏 raised to the power of 𝑛 is equal to 𝑏 raised to
the power of 𝑚 minus 𝑛.
We can use this result to simplify
each division, where we note that 𝑥 and 𝑦 are nonzero since we cannot divide by
zero. We can rewrite 𝑦 as 𝑦 raised to
the first power and apply the quotient rule for exponents on the first term to
obtain two 𝑥 raised to the power of five minus two 𝑦 raised to the power of two
minus one.
We can apply the same process to
the second term to get 𝑘 over two times 𝑥 raised to the power of seven minus two
𝑦 raised to the power of three minus one. We can then evaluate the
expressions in the exponents to obtain the following equation. We can then note that the first
terms on both sides of the equation are identical.
Next, we can see that the second
terms have the same variables raised to the same exponents. Therefore, for the expressions to
be equivalent, the coefficients of these terms must be equal. Hence, 𝑘 over two must be equal to
three, which we can solve to show that 𝑘 equals six.
It is worth noting that since we
want both sides of the equation to be identical, we can verify this answer by
substituting valid values of 𝑥 and 𝑦 into both sides of the equation. For instance, let’s substitute 𝑥
equals one and 𝑦 equals one. If we did this, then all of the
variables become factors of powers of one. So the equation simplifies to give
four plus 𝑘 all over two is equal to two plus three. We can then solve this equation for
𝑘, and we find that 𝑘 equals six.
It is worth noting that this is not
a proof that 𝑘 is equal to six, since we need to show it is true for all valid
values of 𝑥 and 𝑦. And this method only considers one
such pair of values. However, it does help verify that
our answer of 𝑘 equals six is correct.
