Question 01
if x = \mathtt{x=2^{\frac{1}{3}} +2^{\frac{2}{3}}}
Then show that, \mathtt{x^{3} -6x\ =6}
Solution
Finding the value of \mathtt{x^{3}}
\mathtt{x^{3} =\left( 2^{\frac{1}{3}} +2^{\frac{2}{3}}\right)^{3}}
Using the following formula;
\mathtt{(a+b)^{3} =a^{3} +b^{3} +3a^{2} b+3ab^{2}}
Putting the value;
\mathtt{\Longrightarrow \ \left( 2^{\frac{1}{3}} +2^{\frac{2}{3}}\right)^{3}}\\\ \\ \mathtt{\Longrightarrow \ \left( 2^{\frac{1}{3}}\right)^{3} \ +\ \left( 2^{\frac{2}{3}}\right)^{3} +3.2^{\frac{2}{3}} .2^{\frac{2}{3}} +3.2^{\frac{1}{3}} .2^{\frac{4}{3}}}\\\ \\ \mathtt{\Longrightarrow \ 2\ +2^{2} +3.2^{\frac{4}{3}} +3.2^{\frac{5}{3}}}\\\ \\ \mathtt{\Longrightarrow \ 6+3.2\ \left( 2^{\frac{1}{3}} +2^{\frac{2}{3}}\right)}\\\ \\ \mathtt{\Longrightarrow \ 6+6x}
So we got the value of \mathtt{x^{3}} as given below.
\mathtt{x^{3} =\ 6+6x}
Rearranging the expression we get;
\mathtt{x^{3} -6x=\ 6}
Hence Proved.
Question 02
If \mathtt{9^{x+2} =240\ +\ 9^{x}} ;
Find the value of \mathtt{( 8x)^{x}}
Solution
\mathtt{9^{x+2} =240\ +\ 9^{x}}\\\ \\ \mathtt{9^{x} .9^{2} =240+9^{x}}\\\ \\ \mathtt{81.9^{x} -9^{x} =240}\\\ \\ \mathtt{9^{x}( 81-1) \ =\ 240}\\\ \\ \mathtt{9^{x} .80\ =\ 240}\\\ \\ \mathtt{9^{x} =\frac{240}{80}}\\\ \\ \mathtt{9^{x} =\ 3}\\\ \\ \mathtt{3^{2x} =3^{1}}
Since the base are same ,the exponent will also be the same.
2x = 1
x = 1/2
Hence, we got the value of x = 1/2.
Now let’s find the value of \mathtt{( 8x)^{x}}
\mathtt{( 8x)^{x}}\\\ \\ \mathtt{\Longrightarrow \ \left( 8.\frac{1}{2}\right)^{\frac{1}{2}}}\\\ \\ \mathtt{\Longrightarrow \ 4^{\frac{1}{2}}}\\\ \\ \mathtt{\Longrightarrow \ 2}
Hence, 2 is the right answer.
Question 03
Solve and find value of x.
\mathtt{2^{x+1} =8\times 2^{4}}
Solution
\mathtt{2^{x+1} =8\times 2^{4}}\\\ \\ \mathtt{2^{x+1} =2^{3} \times 2^{4}}\\\ \\ \mathtt{2^{x+1} =2^{7}}
Since base are same, the exponent will also be same.
x + 1 = 7
x = 6
Hence, the value of 6 is right answer.
Question 04
Find the value of x.
\mathtt{2^{3x-7} =256}
Solution
\mathtt{2^{3x-7} =256}\\\ \\ \mathtt{2^{3x-7} \ =\ 2^{8}}
Since base are same, exponents will also be the same.
\mathtt{3x\ -\ 7\ =\ 8}\\\ \\ \mathtt{3x\ =\ 8\ +\ 7}\\\ \\ \mathtt{3x\ =\ 15}\\\ \\ \mathtt{x\ =\ \frac{15}{3} =\ 5}
Hence, value of x is 5.
Question 05
From the below expression, find the value of x.
\mathtt{4^{x-1} \times ( 0.5)^{3-2x} =\left(\frac{1}{8}\right)^{x}}
Solution
\mathtt{\left( 2^{2}\right)^{x-1} \times \left(\frac{5}{10}\right)^{3-2x} =\left(\frac{1}{2^{3}}\right)^{x}}\\\ \\ \mathtt{2^{2x-2} \times \left(\frac{1}{2}\right)^{3-2x} =\left( 2^{-3}\right)^{x}}\\\ \\ \mathtt{2^{2x-2} \times \left( 2^{-1}\right)^{3-2x} =( 2)^{-3x}}\\\ \\ \mathtt{2^{2x-2} \times ( 2)^{-3+2x} =( 2)^{-3x}}\\\ \\ \mathtt{2^{2x-2-3+2x} =2^{-3x}}\\\ \\ \mathtt{2^{4x-5} =2^{-3x}}
Since base are same, exponents will also be the same.
\mathtt{4x-5\ =\ -3x}\\\ \\ \mathtt{4x+3x\ =\ 5}\\\ \\ \mathtt{7x\ =\ 5}\\\ \\ \mathtt{x\ =\ \frac{5}{7}}
Hence, the value of x is 5/7.
Next Chapter : Exponents grade 09 problems – set 04