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show that x^{3} − 6x = 6
x^{3 }− 6x = 6
Putting cube on both the sides, we get
As we know, (a+b)3 = a^{3} + b^{3} + 3ab(a+b)
x^{3} = 6 + 6x
x^{3} – 6x = 6
Hence proved
12. Determine (8x)^{x}, if 9^{x+2 }= 240 + 9^{x}.
9^{x+2 }= 240 + 9^{x}
9^{x} .9^{2} = 240 + 9^{x}
Let 9^{x} be y
81y = 240 + y
81y – y = 240
80y = 240
y = 3
Since, y = 3
Then,
9^{x} = 3
3^{2x} = 3
Therefore, x = 1/2
= 2
Therefore (8x)^{x} = 2
13. If 3^{x+1} = 9^{x2}, find the value of 2^{1+x}
3^{x+1} = 9^{x2}
3^{x+1} = 3^{2x4}
x + 1 = 2x – 4
x = 5
Therefore the value of 2^{1+x} = 2^{1+5} = 2^{6} = 64
14. If 3^{4x} = (81)^{1} and (10)^{1/y }= 0.0001, find the value of 2^{x+4y}.
3^{4x} = (81)^{1} and (10)^{1/y} = 0.0001
3^{4x} = (3)^{4}
x = 1
And (10)^{1/y} = 0.0001
(10)^{1/y} = (10)^{−4}
To find the value of 2^{x+4y}, we need to substitute the value of x and y
15. If 5^{3x} = 125 and 10^{y} = 0.001. Find x and y.
5^{3x} = 125 and 10^{y} = 0.001
5^{3x} = 5^{3}
x = 1
Now,
10^{y} = 0.001
10^{y} = 10^{3}
y = 3
Therefore, the value of x = 1 and the value of y = 3
16. Solve the following equations
(i) 3^{x+1} = 2^{7×34}
3^{x+1} = 3^{3}×3^{4}
3^{x+1} = 3^{3+4}
x + 1 = 3 + 4 [By equating exponents]
x + 1 = 7
x = 7 − 1
x = 6
2^{4x} = 2^{3}
4x = 3 (By equating exponents)
x = 3/4
(By equating exponents)
(iii). 3^{x−1}×5^{2y−3 }= 225
3^{x−1}×5^{2y−3 }= 3^{2}×5^{2}
x−1 = 2 [By equating exponents]
x = 3
3^{x−1}×5^{2y−3} = 3^{2}×5^{2}
2y−3 = 2 [By equating exponents]
2y = 5
y = 5/2
3x+3 = 4y+8 —eq1
3(3x+3) = 6+2x+24
9x+9 = 30+2x
7x = 21
x = 21/7
x = 3
Putting value of x in eq2
2x−2−3+2x = −3x [By equating exponents]
4x+3x = 5
7x = 5
x = 5/7
[By equating exponents]
17. If a and b are distinct positive primes such that find x and y
18.If a and b are different positive primes such that
a^{x}b^{y} find x and y
(ii) (a+b)^{−1}(a^{−1}+b^{−1}) = a^{x}b^{y},find x and y
=(ab)^{−1 }= a^{−1}b^{−1 }
By equating exponents
x = −1,y = −1
Therefore x+y+2 = −1−1+2 = 0
19. If 2^{x}×3^{y}×5^{z} = 2160 , find x,y and z. Hence compute the value of 3^{x}×2^{−y}×5^{−z}
20. If 1176 = 2^{a}×3^{b}×7^{c}, find the values of a,b and c. Hence compute the value of 2^{a}×3^{b}×7^{−c} as a fraction
We have to find the value of
= 24/49
21. Simplify
22. Show that
Hence LHS = RHS
23.
(i)
= x^{0} y^{0}
= 1
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