Question

If \[{x^5} - 8 = 159\], what is the approximate value of x?
A) 2.67
B) 2.71
C) 2.78
D) 2.81
E) 2.84

Answer 1

Hint:
We can simplify the equation by taking the constant terms to one side. Then we can take the \[{5^{th}}\] root on both sides. Then we can take logarithms on both sides. After simplification using the properties of logarithms, we can take antilog on both sides by using a logarithmic table. Then we can compare with the options to get the approximate value of x.

Complete step by step solution:
We are given the equation \[{x^5} - 8 = 159\].
We can add 8 on both sides of the equation, so We get
 \[ \Rightarrow {x^5} = 159 + 8\]
 \[ \Rightarrow {x^5} = 167\]
On taking the \[{5^{th}}\] root on both sides, we get
 \[ \Rightarrow x = {\left( {167} \right)^{\dfrac{1}{5}}}\]
Now we can take the logarithm on both sides. So, We get
 \[ \Rightarrow \log x = \log {\left( {167} \right)^{\dfrac{1}{5}}}\]
We know that \[\log {x^a} = a\log x\]. So, we get
 \[ \Rightarrow \log x = \dfrac{1}{5}\log \left( {167} \right)\]
Now we can write $167 = 1.67 \times {10^2}$
 \[ \Rightarrow \log x = \dfrac{1}{5}\log \left( {1.67 \times {{10}^2}} \right)\]
Using properties of log, we can expand the logarithm of the product as
 \[ \Rightarrow \log x = \dfrac{1}{5}\left( {\log 1.67 + \log {{10}^2}} \right)\]
From the logarithmic table, we get $\log 1.67 = 0.222$ and we know that $\log {10^2} = 2$
 \[ \Rightarrow \log x = \dfrac{1}{5}\left( {0.222 + 2} \right)\]
On simplification, We get
 \[ \Rightarrow \log x = \dfrac{{2.222}}{5}\]
 \[ \Rightarrow \log x = 0.4444\]
Now we can take antilogarithm on both sides.
 \[ \Rightarrow x = {10^{0.4444}}\]
Using antilogarithm table, we get
 \[ \Rightarrow x = 2.7822\]
On approximating, we get
 \[ \Rightarrow x = 2.78\]

So, the correct answer is option C which is 2.78

Note:
Alternate solution to this problem is given by,
We are given the equation \[{x^5} - 8 = 159\].
We can add 8 on both sides of the equation, so We get
 \[ \Rightarrow {x^5} = 159 + 8\]
 \[ \Rightarrow {x^5} = 167\]
Now we can check by taking the \[{5^{th}}\] power of each option.
Consider option A, 2.67
 $ \Rightarrow {\left( {2.67} \right)^5} = 135.69$
Consider option B, 2.71
 $ \Rightarrow {\left( {2.71} \right)^5} = 146.16$
Consider option C, 2.78
 $ \Rightarrow {\left( {2.78} \right)^5} = 166.04$
Consider option D, 2.81
 $ \Rightarrow {\left( {2.81} \right)^5} = 175.19$
Consider option E, 2.84
 $ \Rightarrow {\left( {2.84} \right)^5} = 184.75$
From the \[{5^{th}}\] powers of each of the options, we can say that the most approximate solution is obtained when $x = 2.78$
So, the correct answer is option C which is 2.78