
If ${a^x} = {b^y} = {c^z}$and ${b^2} = ac$, then the value of $y$is
1.$\dfrac{{2xz}}{{\left( {x + z} \right)}}$
2. $\dfrac{{xz}}{{2\left( {x - z} \right)}}$
3. $\dfrac{{xz}}{{2\left( {z - x} \right)}}$
4. $\dfrac{{2xz}}{{\left( {x - z} \right)}}$
Answer
146.4k+ views
Hint: In this question, we are given the condition ${a^x} = {b^y} = {c^z}$ Now let these terms be equal to any variable $\left( p \right)$ to calculate the values of $a,b,c$ because we have one more given condition i.e., ${b^2} = ac$ and for this we need the value of $a,b,c$. Now you will have to apply the product rule as you will get the exponentials in multiplication which have the same base. In the last comparison, the exponents of both sides and the value of $y$ will be there.
Formula Used:
Product rule – When we multiply two exponentials with the same base, we simply add the exponents:
${a^p}{a^q} = {a^{p + q}}$
Complete step by step Solution:
Given that,
${a^x} = {b^y} = {c^z}$ and ${b^2} = ac$
Let, ${a^x} = {b^y} = {c^z} = p$
Solve above equality to find the value of $a,b,c$
Therefore, $a = {p^{\dfrac{1}{x}}},b = {p^{\dfrac{1}{y}}},c = {p^{\dfrac{1}{z}}}$
Also given, ${b^2} = ac - - - - - \left( 1 \right)$
Put the values of $a,b,c$in equation (1)
It implies that,
${\left( {{p^{\dfrac{1}{y}}}} \right)^2} = {p^{\dfrac{1}{x}}}{p^{\dfrac{1}{z}}}$
Solve the right-side by adding the exponents as base of the exponentials are same.
${p^{\dfrac{2}{y}}} = {p^{\dfrac{1}{x} + \dfrac{1}{z}}}$
${p^{\dfrac{2}{y}}} = {p^{\dfrac{{z + x}}{{xz}}}}$
Compare the exponentials of both sides as the base is same
$ \Rightarrow \dfrac{2}{y} = \dfrac{{z + x}}{{xz}}$
Now cross multiplying the term $2$ to the right side,
$\dfrac{1}{y} = \dfrac{{x + z}}{{2xz}}$
Reciprocal both sides,
$y = \dfrac{{2xz}}{{x + z}}$
Hence, the correct option is 1.
Note: The key concept involved in solving this problem is a good knowledge of power and exponent. Students must remember that power is defined as a base number multiplied by its exponent, where the base number is the factor multiplied by itself and the exponent represents how many times the same base number is multiplied.
Formula Used:
Product rule – When we multiply two exponentials with the same base, we simply add the exponents:
${a^p}{a^q} = {a^{p + q}}$
Complete step by step Solution:
Given that,
${a^x} = {b^y} = {c^z}$ and ${b^2} = ac$
Let, ${a^x} = {b^y} = {c^z} = p$
Solve above equality to find the value of $a,b,c$
Therefore, $a = {p^{\dfrac{1}{x}}},b = {p^{\dfrac{1}{y}}},c = {p^{\dfrac{1}{z}}}$
Also given, ${b^2} = ac - - - - - \left( 1 \right)$
Put the values of $a,b,c$in equation (1)
It implies that,
${\left( {{p^{\dfrac{1}{y}}}} \right)^2} = {p^{\dfrac{1}{x}}}{p^{\dfrac{1}{z}}}$
Solve the right-side by adding the exponents as base of the exponentials are same.
${p^{\dfrac{2}{y}}} = {p^{\dfrac{1}{x} + \dfrac{1}{z}}}$
${p^{\dfrac{2}{y}}} = {p^{\dfrac{{z + x}}{{xz}}}}$
Compare the exponentials of both sides as the base is same
$ \Rightarrow \dfrac{2}{y} = \dfrac{{z + x}}{{xz}}$
Now cross multiplying the term $2$ to the right side,
$\dfrac{1}{y} = \dfrac{{x + z}}{{2xz}}$
Reciprocal both sides,
$y = \dfrac{{2xz}}{{x + z}}$
Hence, the correct option is 1.
Note: The key concept involved in solving this problem is a good knowledge of power and exponent. Students must remember that power is defined as a base number multiplied by its exponent, where the base number is the factor multiplied by itself and the exponent represents how many times the same base number is multiplied.
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