
If \[{x^{3a}} = {y^{2b}} = {z^{4c}} = xyz\], then find the value of \[3ab + 4bc + 6ca\]
Answer
587.1k+ views
Hint: If two powers have the same base then we can multiply the powers. When we multiply two same variables we add their exponents that is powers which can be represented by the following formula ${a^m}{a^n} = {a^{m + n}}$
Complete step-by-step answer:
It is given in the question that \[{x^{3a}} = {y^{2b}} = {z^{4c}} = xyz\]
Now let us write y and z in terms of x using given relation,
Initially let us consider \[{x^{3a}} = {y^{2b}}\], on simplifying this we get,
\[y = {x^{\dfrac{{3a}}{{2b}}}}\], and
Let us now consider \[{x^{3a}} = {z^{4c}}\], this equality from the given relation therefore we get,
\[z = {x^{\dfrac{{3a}}{{4c}}}}\]
Now we shall use the values of y and z in xyz we get,
\[xyz = x \times {x^{\dfrac{{3a}}{{2b}}}} \times {x^{\dfrac{{3a}}{{4c}}}}\]
Here we can use the identity ${a^m}{a^n} = {a^{m + n}}$ in the variable x, therefore we get,
\[xyz = {x^{1 + \dfrac{{3a}}{{2b}} + \dfrac{{3a}}{{4c}}}}\].........(1)
Mark it as equation (1)
From the given relation \[{x^{3a}} = {y^{2b}} = {z^{4c}} = xyz\]
Consider \[xyz = {x^{3a}}\].........(2)
Mark it as equation (2)
Let us now equate the equation (1) and (2) as the left hand side of both the equation remains the same.
On equating (1) and (2) we get,
\[{x^{3a}} = {x^{1 + \dfrac{{3a}}{{2b}} + \dfrac{{3a}}{{4c}}}}\]
Since the variable in both side is the same it is clear that the power of the variable remains the same, hence we get,
\[3a = 1 + \dfrac{{3a}}{{2b}} + \dfrac{{3a}}{{4c}}\]
Let us now solve the equation on the right hand side so that it can be further simplified.
\[3a = \dfrac{{4bc + 6ac + 3ab}}{{4bc}}\]
Now let us multiply by 4bc in both sides of the equation therefore we get,
\[3ab + 4bc + 6ca = 12abc\]
Whereas the equation on the left hand side is what we have to find
Hence we have found the value of \[3ab + 4bc + 6ca\].
The value of \[3ab + 4bc + 6ca\] is \[12abc\]
Note: If we equate two same variables it is true that the power will be the same on both sides this is given by the identity ${x^m} = {x^n} \Rightarrow m = n$. This is one of the important identities used in the problem.
Complete step-by-step answer:
It is given in the question that \[{x^{3a}} = {y^{2b}} = {z^{4c}} = xyz\]
Now let us write y and z in terms of x using given relation,
Initially let us consider \[{x^{3a}} = {y^{2b}}\], on simplifying this we get,
\[y = {x^{\dfrac{{3a}}{{2b}}}}\], and
Let us now consider \[{x^{3a}} = {z^{4c}}\], this equality from the given relation therefore we get,
\[z = {x^{\dfrac{{3a}}{{4c}}}}\]
Now we shall use the values of y and z in xyz we get,
\[xyz = x \times {x^{\dfrac{{3a}}{{2b}}}} \times {x^{\dfrac{{3a}}{{4c}}}}\]
Here we can use the identity ${a^m}{a^n} = {a^{m + n}}$ in the variable x, therefore we get,
\[xyz = {x^{1 + \dfrac{{3a}}{{2b}} + \dfrac{{3a}}{{4c}}}}\].........(1)
Mark it as equation (1)
From the given relation \[{x^{3a}} = {y^{2b}} = {z^{4c}} = xyz\]
Consider \[xyz = {x^{3a}}\].........(2)
Mark it as equation (2)
Let us now equate the equation (1) and (2) as the left hand side of both the equation remains the same.
On equating (1) and (2) we get,
\[{x^{3a}} = {x^{1 + \dfrac{{3a}}{{2b}} + \dfrac{{3a}}{{4c}}}}\]
Since the variable in both side is the same it is clear that the power of the variable remains the same, hence we get,
\[3a = 1 + \dfrac{{3a}}{{2b}} + \dfrac{{3a}}{{4c}}\]
Let us now solve the equation on the right hand side so that it can be further simplified.
\[3a = \dfrac{{4bc + 6ac + 3ab}}{{4bc}}\]
Now let us multiply by 4bc in both sides of the equation therefore we get,
\[3ab + 4bc + 6ca = 12abc\]
Whereas the equation on the left hand side is what we have to find
Hence we have found the value of \[3ab + 4bc + 6ca\].
The value of \[3ab + 4bc + 6ca\] is \[12abc\]
Note: If we equate two same variables it is true that the power will be the same on both sides this is given by the identity ${x^m} = {x^n} \Rightarrow m = n$. This is one of the important identities used in the problem.
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