log base(7-4√3)^(7+4√3)
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Step-by-step explanation:
The value of a=97 and b=-56.
Step-by-step explanation:
Given : \frac{7-4\sqrt{3}}{7+4\sqrt{3}}=a+b\sqrt{3}
7+4
3
7−4
3
=a+b
3
To find : The value of a and b?
Solution :
We solve the given expression LHS by rationalizing,=
7+4
3
7−4
3
×
7+4
3
7−4
3
Applying property, (a-b)(a+b)=a^2-b^2(a−b)(a+b)=a
2
−b
2
=\frac{(7-4\sqrt{3})^2}{7^2-(4\sqrt{3})^2}=
7
2
−(4
3
)
2
(7−4
3
)
2
=\frac{49+48-56\sqrt3}{49-48}=
49−48
49+48−56
3
=\frac{49+48-56\sqrt3}{1}=
1
49+48−56
3
=97-56\sqrt3=97−56
3
On comparing with RHS,
a+b\sqrt{3}=97-56\sqrt3a+b
3
=97−56
3
a=97 and b=-56
Therefore, The value of a=97 and b=-56.
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Answer:
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