
Solve the following equation for the value of x:
\[3 \times {x^{{{\log }_5}2}} + {2^{{{\log }_5}x}} = 64\]
Answer
615.9k+ views
Hint: To find the value of x in the given equation we have made all the powers of the equation. So, we will use logarithmic identity \[{a^{{{\log }_b}c}} = {c^{{{\log }_c}b}}\]. To make the powers equal to the equation. And then we can easily solve the equation by comparing the powers.
Complete step-by-step answer:
As we know that the logarithmic equation given to us is,
\[3 \times {x^{{{\log }_5}2}} + {2^{{{\log }_5}x}} = 64\] (1)
Now to find the value of x from the above equation. We must have to make the powers of x and 2 in the above equation the same.
So, let us use logarithmic identity to do so.
As we know that if a, b and c are any numbers then then \[{a^{{{\log }_b}c}}\] can also be written as \[{c^{{{\log }_c}b}}\] .
So, in the equation 1, \[{x^{{{\log }_5}2}}\] can also be written as \[{2^{{{\log }_5}x}}\].
So, now equation 1 becomes,
\[3 \times {2^{{{\log }_5}x}} + {2^{{{\log }_5}x}} = 64\] …...(2)
Now to find the value of x let us take \[{2^{{{\log }_5}x}}\] common from the LHS of equation 2.
So, equation 2 becomes,
\[{2^{{{\log }_5}x}}\left( {3 + 1} \right) = 4 \times {2^{{{\log }_5}x}} = 64\]
Now dividing both sides of the above equation by 4. We get,
\[{2^{{{\log }_5}x}} = 16\]
As we know that 16 can also be written as \[{2^4}\].
\[{2^{{{\log }_5}x}} = {2^4}\]
And if \[{a^b} = {a^c}\], then a = c.
So, \[{\log _5}x\] = 4 …..(3)
And as we know according to the logarithmic property if \[{\log _a}b = c\] then the value of b will be calculated as \[b = {a^c}\].
So, now we find the value of x from equation 3 using logarithmic property.
So, \[x = {5^4} = 625\]
Hence, the value of x will be equal to 625.
Note: Whenever we come up with this type of problem then first, we had to make the equation such that we can take the logarithmic part of the equation common using identity which states that \[{a^{{{\log }_b}c}} = {c^{{{\log }_c}b}}\] then we had to solve the rest of the equation and after that we can compare the powers of LHS and RHS to get the value of logarithmic part and after that we can the value of x using the identity which states that if \[{\log _a}b = c\] then \[b = {a^c}\].
Complete step-by-step answer:
As we know that the logarithmic equation given to us is,
\[3 \times {x^{{{\log }_5}2}} + {2^{{{\log }_5}x}} = 64\] (1)
Now to find the value of x from the above equation. We must have to make the powers of x and 2 in the above equation the same.
So, let us use logarithmic identity to do so.
As we know that if a, b and c are any numbers then then \[{a^{{{\log }_b}c}}\] can also be written as \[{c^{{{\log }_c}b}}\] .
So, in the equation 1, \[{x^{{{\log }_5}2}}\] can also be written as \[{2^{{{\log }_5}x}}\].
So, now equation 1 becomes,
\[3 \times {2^{{{\log }_5}x}} + {2^{{{\log }_5}x}} = 64\] …...(2)
Now to find the value of x let us take \[{2^{{{\log }_5}x}}\] common from the LHS of equation 2.
So, equation 2 becomes,
\[{2^{{{\log }_5}x}}\left( {3 + 1} \right) = 4 \times {2^{{{\log }_5}x}} = 64\]
Now dividing both sides of the above equation by 4. We get,
\[{2^{{{\log }_5}x}} = 16\]
As we know that 16 can also be written as \[{2^4}\].
\[{2^{{{\log }_5}x}} = {2^4}\]
And if \[{a^b} = {a^c}\], then a = c.
So, \[{\log _5}x\] = 4 …..(3)
And as we know according to the logarithmic property if \[{\log _a}b = c\] then the value of b will be calculated as \[b = {a^c}\].
So, now we find the value of x from equation 3 using logarithmic property.
So, \[x = {5^4} = 625\]
Hence, the value of x will be equal to 625.
Note: Whenever we come up with this type of problem then first, we had to make the equation such that we can take the logarithmic part of the equation common using identity which states that \[{a^{{{\log }_b}c}} = {c^{{{\log }_c}b}}\] then we had to solve the rest of the equation and after that we can compare the powers of LHS and RHS to get the value of logarithmic part and after that we can the value of x using the identity which states that if \[{\log _a}b = c\] then \[b = {a^c}\].
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