
The value of ${7^{{{\log }_9}5}} - {5^{{{\log }_9}7}}$is
$\left( a \right){\text{ 0}}$
$\left( b \right){\text{ 1}}$
$\left( c \right){\text{ 10}}$
$\left( d \right){\text{ lo}}{{\text{g}}_9}5$
Answer
559.2k+ views
Hint: So for solving this question, we must remember the properties associated with this problem which is ${a^{{{\log }_c}b}} = {b^{{{\log }_c}a}}$. Here the base remains the same and the power gets interchanged. So by using this property we can easily solve this.
Formula used:
One of the properties of logarithmic function will be-
${a^{{{\log }_c}b}} = {b^{{{\log }_c}a}}$
Here,
$a$& $b$ gets interchanged while $c$ will be the base which will be constant throughout.
Complete step-by-step answer:
We all know the property of log which is-
${a^{{{\log }_c}b}} = {b^{{{\log }_c}a}}$
Now on implementing this property to the question, we can write it as
$ \Rightarrow {7^{{{\log }_9}5}} = {5^{{{\log }_9}7}}$
From above we can see that the base is the same in both and only $a$& $b$ is changing. So we can say it is validating this property.
So from the question, the equation will become
$ \Rightarrow {7^{{{\log }_9}5}} - {5^{{{\log }_9}7}}$
Now since both are equal to each other so we can write it as,
$ \Rightarrow {5^{{{\log }_9}7}} - {5^{{{\log }_9}7}}$
Since both are the same, so it will cancel each other and we get
$ \Rightarrow 0$
Therefore, the option $\left( a \right)$ will be correct.
Additional information: Logarithms reduce the strength of arithmetic operations, dropping multiplication down to addition (and reducing exponentiation down to multiplication). And also we should keep in mind about the only thing we need to keep in mind is what we are doing. Just remember the exponent law.
Note: It can also be solved by using the formula ${\log _n}m = \dfrac{{\log m}}{{\log n}}$. So for solving it we will first name both the log and then solve for them independently. After solving both the logarithmic we will see that both are equal to each other and hence on subtracting we will get the result. So in this way also we can solve this problem.
Formula used:
One of the properties of logarithmic function will be-
${a^{{{\log }_c}b}} = {b^{{{\log }_c}a}}$
Here,
$a$& $b$ gets interchanged while $c$ will be the base which will be constant throughout.
Complete step-by-step answer:
We all know the property of log which is-
${a^{{{\log }_c}b}} = {b^{{{\log }_c}a}}$
Now on implementing this property to the question, we can write it as
$ \Rightarrow {7^{{{\log }_9}5}} = {5^{{{\log }_9}7}}$
From above we can see that the base is the same in both and only $a$& $b$ is changing. So we can say it is validating this property.
So from the question, the equation will become
$ \Rightarrow {7^{{{\log }_9}5}} - {5^{{{\log }_9}7}}$
Now since both are equal to each other so we can write it as,
$ \Rightarrow {5^{{{\log }_9}7}} - {5^{{{\log }_9}7}}$
Since both are the same, so it will cancel each other and we get
$ \Rightarrow 0$
Therefore, the option $\left( a \right)$ will be correct.
Additional information: Logarithms reduce the strength of arithmetic operations, dropping multiplication down to addition (and reducing exponentiation down to multiplication). And also we should keep in mind about the only thing we need to keep in mind is what we are doing. Just remember the exponent law.
Note: It can also be solved by using the formula ${\log _n}m = \dfrac{{\log m}}{{\log n}}$. So for solving it we will first name both the log and then solve for them independently. After solving both the logarithmic we will see that both are equal to each other and hence on subtracting we will get the result. So in this way also we can solve this problem.
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