How do you solve ${\log _5}x = \log 80$?
Answer
615k+ views
Hint: We will, first of all, use the identity that: If ${\log _b}a = x$, then $a = {b^x}$. Then we will make use of the fact that log(a.b) = log a + log b. Then we will use the fact that log 10 = 1.
Complete step by step solution:
We are given that we are required to solve ${\log _5}x = \log 80$.
Since, we know that we have a fact given by the following expression:-
If ${\log _b}a = x$, then $a = {b^x}$
Replacing b by 5, x by log 80 and a by x, we will then obtain the following equation with us:-
$ \Rightarrow x = {5^{\log 80}}$
Since we know that 80 can be written as multiple of 8 and 10.
Therefore, we can write the above mentioned expression as follows:-
$ \Rightarrow x = {5^{\log (8 \times 10)}}$
Now, we will use the fact that $\log a.b = \log a + \log b$. Then, just by replacing a by 8 and b by10, we will then obtain the following equation with us:-
$ \Rightarrow \log (8 \times 10) = \log 8 + \log 10$
Putting this in the expression given to us, we will then obtain the following equation with us:-
$ \Rightarrow x = {5^{\log 8 + \log 10}}$
Since, we know that we have a fact given by the following expression:-
$ \Rightarrow $ log 10 = 1
Putting this in the given expression, we will then obtain the following expression with us:-
$ \Rightarrow x = {5^{\log 8 + 1}}$
Note: The students must commit to memory the following facts we used in the solution given above:-
1. If ${\log _b}a = x$, then $a = {b^x}$.
2. $\log a.b = \log a + \log b$
3. log 10 = 1
This is true for all positive real numbers or in some cases real numbers as well.
The students must also keep in mind that the logarithm of 0 or any negative terms are also not defined. The logarithmic is only defined for positive values. The power may be positive, negative or zero.
Therefore, in the first formula stated above in the note, a can never be equal to 0 or any negative number which can be expressed as: a > 0 but b can take any real value.
In the second formula stated in the note above, a and b can never be equal to 0 or any negative number which can be expressed as a, b > 0
Complete step by step solution:
We are given that we are required to solve ${\log _5}x = \log 80$.
Since, we know that we have a fact given by the following expression:-
If ${\log _b}a = x$, then $a = {b^x}$
Replacing b by 5, x by log 80 and a by x, we will then obtain the following equation with us:-
$ \Rightarrow x = {5^{\log 80}}$
Since we know that 80 can be written as multiple of 8 and 10.
Therefore, we can write the above mentioned expression as follows:-
$ \Rightarrow x = {5^{\log (8 \times 10)}}$
Now, we will use the fact that $\log a.b = \log a + \log b$. Then, just by replacing a by 8 and b by10, we will then obtain the following equation with us:-
$ \Rightarrow \log (8 \times 10) = \log 8 + \log 10$
Putting this in the expression given to us, we will then obtain the following equation with us:-
$ \Rightarrow x = {5^{\log 8 + \log 10}}$
Since, we know that we have a fact given by the following expression:-
$ \Rightarrow $ log 10 = 1
Putting this in the given expression, we will then obtain the following expression with us:-
$ \Rightarrow x = {5^{\log 8 + 1}}$
Note: The students must commit to memory the following facts we used in the solution given above:-
1. If ${\log _b}a = x$, then $a = {b^x}$.
2. $\log a.b = \log a + \log b$
3. log 10 = 1
This is true for all positive real numbers or in some cases real numbers as well.
The students must also keep in mind that the logarithm of 0 or any negative terms are also not defined. The logarithmic is only defined for positive values. The power may be positive, negative or zero.
Therefore, in the first formula stated above in the note, a can never be equal to 0 or any negative number which can be expressed as: a > 0 but b can take any real value.
In the second formula stated in the note above, a and b can never be equal to 0 or any negative number which can be expressed as a, b > 0
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