Question

How do you approximate ${{\log }_{5}}50$ given ${{\log }_{5}}2=0.4307$ and ${{\log }_{5}}3=0.6826?$

Answer 1

Hint: We use the properties of logarithm to solve this problem. An important rule we will be using is ${{\log }_{n}}a=\dfrac{\log a}{\log n}.$ Another rule we will be using is $\log {{a}^{n}}=n\log a.$ Also, we will use ${{\log }_{a}}mn={{\log }_{a}}m+{{\log }_{a}}n.$

Complete step by step solution:
Let us consider the logarithmic value we are asked to find, ${{\log }_{5}}50$
Also, we are given with the values, ${{\log }_{5}}2=0.4307$ and ${{\log }_{5}}3=0.6826.$
We can solve this problem without using the value ${{\log }_{5}}3=0.6826.$
That is, we can find the value of ${{\log }_{5}}50$ only with the help of ${{\log }_{5}}2.$
We know that $50=2\times 25.$
So, we can write ${{\log }_{5}}50={{\log }_{5}}2\times 25.$
Now, we are using the property given as ${{\log }_{a}}mn={{\log }_{a}}m+{{\log }_{a}}n.$
We are going to apply this in the above written equation.
Then we get the following,
$\Rightarrow {{\log }_{5}}50={{\log }_{5}}2\times 25={{\log }_{5}}2+{{\log }_{5}}25.$
Now we will apply the value which we are given with in the above obtained equation.
That is, ${{\log }_{5}}2=0.4307.$
 Then we will get,
$\Rightarrow {{\log }_{5}}50={{\log }_{5}}2\times 25=0.4307+{{\log }_{5}}25.$
Now, in the above equation, we do not have what the value of ${{\log }_{5}}25$ is.
We know that $25={{5}^{2}}.$
So, we will get, ${{\log }_{5}}25={{\log }_{5}}{{5}^{2}}.$
Let us apply this in our equation, we get
$\Rightarrow {{\log }_{5}}50={{\log }_{5}}2\times 25=0.4307+{{\log }_{5}}{{5}^{2}}.$
Now, we will use the property of logarithm ${{\log }_{n}}{{a}^{m}}=m{{\log }_{n}}a.$
When we use this property, we will get ${{\log }_{5}}{{5}^{2}}=2{{\log }_{5}}5.$
Let us substitute this value in the equation we obtained.
Then we will get,
$\Rightarrow {{\log }_{5}}50=0.4307+2{{\log }_{5}}5.$
And we will use a rule given as ${{\log }_{m}}n=\dfrac{\log n}{\log m}$ in the equation.
Thus, we will get ${{\log }_{5}}5=\dfrac{\log 5}{\log 5}$
Now, we are going to apply this in our equation.
We will get,
$\Rightarrow {{\log }_{5}}50=0.4307+\dfrac{2\log 5}{\log 5}.$
We are cancelling $\log 5,$
$\Rightarrow {{\log }_{5}}50=0.4307+2.$
So,
$\Rightarrow {{\log }_{5}}50=2.4307.$
Hence, the actual value of ${{\log }_{5}}50=2.4307.$

Note: What we found is the exact value of ${{\log }_{5}}50.$ By using both the logarithmic values given we can approximate ${{\log }_{5}}50$ to $2.4054.$
Consider $50\to 3\times 16=3\times {{2}^{4}}=48.$
So, we get ${{\log }_{5}}50\approx {{\log }_{5}}3+{{\log }_{5}}{{2}^{4}}={{\log }_{5}}3+4{{\log }_{5}}2.$
Applying the values, $\Rightarrow {{\log }_{5}}50\approx {{\log }_{5}}3+4{{\log }_{5}}2=0.6826+4\times 0.4307=0.6826+1.7228=2.4054.$
The error is $2.4307-2.4054=0.0253.$