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junior college 1 | H1 Maths
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Date Posted: 4 years ago
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i)
k^(ln x) = x^(ln k) , k > 0
x > 0 for k^(ln x) to be defined, since ln x is only defined for x > 0
Taking ln on both sides,
ln k^(ln x) = ln x^(ln k)
ln x (ln k) = ln k (ln x)
ln x = ln x
(Dividing both sides by constant, ln k )
Since LHS = RHS,
This means as long as x > 0, x ∈ R such that ln x is defined, the equation will be satisfied since LHS would equal RHS.
k^(ln x) = x^(ln k) , k > 0
x > 0 for k^(ln x) to be defined, since ln x is only defined for x > 0
Taking ln on both sides,
ln k^(ln x) = ln x^(ln k)
ln x (ln k) = ln k (ln x)
ln x = ln x
(Dividing both sides by constant, ln k )
Since LHS = RHS,
This means as long as x > 0, x ∈ R such that ln x is defined, the equation will be satisfied since LHS would equal RHS.
ii)
(√e)^(ln x) = 26 - 7x
Using the equality in i),
x^(ln √e) = 26 - 7x
x^(ln e^½) = 26 - 7x
x^(½ln e) = 26 - 7x
x¹/² = 26 - 7x
x = (26 - 7x)²
x = 676 - 364x + 49x²
49x² - 365x + 676 = 0
(49x - 169)(x - 4) = 0
49x = 169
x = 169/49 = 3 22/49
Or
x = 4
(rejected as when x = 4,
(√e)^(ln4) = 4¹/² = 2
but 26 - 7x = 26 - 7(4) = -2
So (√e)^(ln x) ≠ 26 - 7x when x = 4)
(√e)^(ln x) = 26 - 7x
Using the equality in i),
x^(ln √e) = 26 - 7x
x^(ln e^½) = 26 - 7x
x^(½ln e) = 26 - 7x
x¹/² = 26 - 7x
x = (26 - 7x)²
x = 676 - 364x + 49x²
49x² - 365x + 676 = 0
(49x - 169)(x - 4) = 0
49x = 169
x = 169/49 = 3 22/49
Or
x = 4
(rejected as when x = 4,
(√e)^(ln4) = 4¹/² = 2
but 26 - 7x = 26 - 7(4) = -2
So (√e)^(ln x) ≠ 26 - 7x when x = 4)
Alternatively, without using the inequality,
(√e)^(ln x) = e^(½ ln x) = 26 - 7x
(e^(½ln x))² = (26 - 7x)²
e^(½ln x · 2) = 676 - 364x + 49x²
e^(ln x) = 676 - 364x + 49x²
x = 676 - 364x + 49x²
Solve accordingly.
(√e)^(ln x) = e^(½ ln x) = 26 - 7x
(e^(½ln x))² = (26 - 7x)²
e^(½ln x · 2) = 676 - 364x + 49x²
e^(ln x) = 676 - 364x + 49x²
x = 676 - 364x + 49x²
Solve accordingly.
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My proof
Date Posted:
4 years ago
For b, let k = sqrt(e). Then using the equality at a, LHS of b can be reduced to x^(1/2). Then the equation of b can be solved easily by squaring both side and then solving quadratic equation.
For part(i), how does ln(k) . ln(x) = ln(x) . ln(k) prove x>0?
In both sides of the equation for a)
ln k and ln x appear at both sides of the equation. Since ln K is a constant, it follows that any real positive x will satisfy the equation.
Thanks!
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