Question 1209558: a) If 7^sin²x - 7^cos²x = 8
find the value of x.
b) Let m be a given real number, Find all complex numbers x such that,
([x/(x+1)]² + [x/(x-1)]² = m² + m
c) ab = ½, bc = ⅓, ac = 1/6,
find (1/a²) + (1/b²) + (1/c²)
d) 2^x - 3^y = 5
2^(x+2) + 3^(y+2) = 59.
find xy
e) find m,
9^4^m = 4^9^m
f) (7^log₈x)×(x^log₉x) = 3969,
find x.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve these problems:
**a) 7^(sin²x) - 7^(cos²x) = 8**
1. Use the identity cos²x = 1 - sin²x:
7^(sin²x) - 7^(1-sin²x) = 8
2. Rewrite the second term:
7^(sin²x) - 7 / 7^(sin²x) = 8
3. Let y = 7^(sin²x):
y - 7/y = 8
4. Multiply by y:
y² - 7 = 8y
5. Rearrange into a quadratic equation:
y² - 8y - 7 = 0
6. Factor:
(y - 7)(y + 1) = 0
7. Solve for y:
y = 7 or y = -1
Since 7^(sin²x) must be positive, y = -1 is not a valid solution. So, y = 7.
8. Substitute back:
7^(sin²x) = 7¹
sin²x = 1
9. Solve for x:
sin x = ±1
x = (2n + 1)π/2, where n is an integer.
**b) [x/(x+1)]² + [x/(x-1)]² = m² + m**
1. Simplify the left side:
x²/(x²+2x+1) + x²/(x²-2x+1) = m² + m
2. Find a common denominator and combine the fractions:
[x²(x²-2x+1) + x²(x²+2x+1)] / [(x²+2x+1)(x²-2x+1)] = m² + m
[x⁴-2x³+x² + x⁴+2x³+x²] / (x⁴-2x²+1) = m² + m
(2x⁴ + 2x²) / (x⁴ - 2x² + 1) = m² + m
2x²(x² + 1) / (x² - 1)² = m(m + 1)
This equation is a bit complex. Without a specific value for 'm', it's difficult to simplify further to directly solve for x. The best approach would be to substitute a given value for m and then try to solve for x.
**c) ab = ½, bc = ⅓, ac = 1/6**
1. Multiply the three equations together:
(ab)(bc)(ac) = (1/2)(1/3)(1/6)
a²b²c² = 1/36
2. Take the square root of both sides:
abc = ±1/6
3. Divide (abc = ±1/6) by each of the original equations to get the reciprocals:
1/c = (abc)/(ab) = (±1/6)/(1/2) = ±1/3, so c² = 9
1/a = (abc)/(bc) = (±1/6)/(1/3) = ±1/2, so a² = 4
1/b = (abc)/(ac) = (±1/6)/(1/6) = ±1, so b² = 1
4. Calculate the sum of the reciprocals squared:
(1/a²) + (1/b²) + (1/c²) = 1/4 + 1 + 9 = 10.25 = 41/4
**d) 2^x - 3^y = 5 and 2^(x+2) + 3^(y+2) = 59**
1. Rewrite the second equation:
4 * 2^x + 9 * 3^y = 59
2. Let u = 2^x and v = 3^y. The system becomes:
u - v = 5
4u + 9v = 59
3. Solve for u and v: From the first equation, u = v + 5. Substituting this into the second equation:
4(v + 5) + 9v = 59
13v + 20 = 59
13v = 39
v = 3
Then, u = v + 5 = 3 + 5 = 8.
4. Substitute back to find x and y:
2^x = 8 = 2³ => x = 3
3^y = 3 = 3¹ => y = 1
5. Find xy:
xy = 3 * 1 = 3
**e) 9^(4^m) = 4^(9^m)**
1. Take the logarithm of both sides (any base, but let's use the natural log):
ln(9^(4^m)) = ln(4^(9^m))
2. Use the logarithm power rule:
4^m * ln(9) = 9^m * ln(4)
3. Rewrite ln(9) as 2ln(3) and ln(4) as 2ln(2):
4^m * 2ln(3) = 9^m * 2ln(2)
4. Simplify:
4^m * ln(3) = 9^m * ln(2)
5. Rearrange:
4^m / 9^m = ln(2) / ln(3)
(2²/3²)^m = ln(2) / ln(3)
(2/3)^(2m) = log₃2
2m * ln(2/3) = ln(log₃2)
m = ln(log₃2) / (2 * ln(2/3))
**f) (7^(log₈x)) * (x^(log₉x)) = 3969**
1. Take the logarithm base 8 of both sides:
log₈[(7^(log₈x)) * (x^(log₉x))] = log₈3969
2. Use logarithm properties:
log₈(7^(log₈x)) + log₈(x^(log₉x)) = log₈3969
(log₈x)(log₈7) + (log₉x)(log₈x) = log₈3969
3. Let y = log₈x:
y * log₈7 + (log₉x) * y = log₈3969
y(log₈7 + log₉x) = log₈3969
We also know that 3969 = 63² = 9*7*9*7 = 3⁴ * 7²
log₈3969 = log₈(3⁴ * 7²) = 4log₈3 + 2log₈7
y(log₈7 + log₉x) = 4log₈3 + 2log₈7
This equation is still quite complex. It's likely there's a clever substitution or manipulation I'm missing to simplify it further and solve directly for x. Numerical methods or approximations might be necessary.
Answer by ikleyn(52778)  (Show Source):
You can put this solution on YOUR website! .
Part (a)
Let's consider part (a) first, with equation
7^(sin^2(x)) - 7^(cos^2(x)) = 8.
Due to OBVIOUS reasons, this equation has no real solutions.
Indeed, sin^2(x) has the maximum value of 1; hence, 7^(sin^2(x)) has maximum value of 7.
Next, from 7^(sin^2(x)), the equation subtracts 7^(cos^2(x)), which is non-negative
real number, so the left side of the equation CAN NOT be greater than 7.
A fortiori, it can not be 8.
At this point, my solution is complete, and the
ANSWER is: this given equation has no solution/solutions in real numbers.
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The solution by the @CPhill to this problem has a FATAL ERROR:
it incorrectly factors y^2 - 8y - 7 = 0 as (y-7)*(y+1) = 0,
which leads him to further wrong conclusions.
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This example/case tells me that @CPhill
- does not look at his solutions;
- does not read his solutions;
- does not check them;
- does not think about them;
- does not care about the correctness of his solutions,
and, in general, does not understand, what he is doing and what he posts
to this forum, in particular, and to the outer world, in general.
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Part (c)
In the post by @CPhill, the solution for part (c) is INCORRECT.
The correct solution is given at this forum under this link
https://www.algebra.com/algebra/homework/playground/test.faq.question.1209563.html
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In future, do not pack more than one problem per post.
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