If 4^x+3^x=5^x, then the value of X=?
If 4^x+3^x=5^x, then the value of
X=?
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To solve for x in the equation 4^x + 3^x = 5^x, we can use logarithms. Taking the logarithm of both sides of the equation with base 10, we get:
xlog10(4) + xlog10(3) = x*log10(5)
Dividing both sides by x, we get:
log10(4) + log10(3) = log10(5)
Using a calculator or logarithm tables, we can find that:
log10(4) ≈ 0.60206
log10(3) ≈ 0.47712
log10(5) ≈ 0.69897
Substituting these values into the equation, we get:
0.60206 + 0.47712 ≈ 1.07918 ≈ 0.69897
Therefore, x ≈ 1.07918.
Is the method of solving correct? This is not correct. Because You can't take the logarithm of numbers added separately.
Cross check
4² + 3² = 25 = 5²
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1. Using logarithmic identity method:
If 4^x+3^x=5^x, then the value of
X=?........
The equation 4^x + 3^x = 5^x can be solved by taking the natural logarithm of both sides. We have:
ln(4^x + 3^x) = ln(5^x)
Using the logarithmic identity ln(a^b) = b*ln(a), we can simplify the left-hand side:
xln(4) + xln(3) = x*ln(5)
Dividing both sides by x gives:
ln(4) + ln(3) = ln(5)
Taking the exponential of both sides gives:
4*3 = 5
which is clearly false. Therefore, there is no value of x that satisfies the equation 4^x + 3^x = 5^x.
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Using Binomial method:
If 4^x+3^x=5^x, then the value of
X=?........
Use binomial method
We can use the binomial theorem to expand the terms on the left-hand side of the equation:
(4+3)^x = 5^x
Simplifying the left-hand side using the binomial theorem, we get:
4^x + 3^x + (terms containing higher powers of 4 and 3)
Since the binomial expansion of (4+3)^x contains only positive terms, we know that the terms containing higher powers of 4 and 3 are positive. Therefore, we can write:
4^x + 3^x < (4+3)^x = 7^x
Substituting this inequality into the original equation, we get:
4^x + 3^x < 5^x
This inequality is true for all x, since the left-hand side is less than the right-hand side. Therefore, there is no value of x that satisfies the equation 4^x + 3^x = 5^x.
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Using Newton Ralph method:
If 4^x+3^x=5^x, then the value of
X=?........
Newton Ralph method
The equation 4^x + 3^x = 5^x can be solved using numerical methods such as the Newton-Raphson method, which is an iterative method for finding the roots of a function. We can rewrite the equation as:
f(x) = 4^x + 3^x - 5^x = 0
The derivative of f(x) with respect to x is:
f'(x) = 4^x * ln(4) + 3^x * ln(3) - 5^x * ln(5)
The Newton-Raphson method starts with an initial guess x_0 and iteratively refines the estimate using the formula:
x_{n+1} = x_n - f(x_n) / f'(x_n)
We can use a calculator or a computer program to implement this iterative formula. For example, using an initial guess of x_0 = 1, we obtain the following values:
x_1 = 1.184666
x_2 = 1.175669
x_3 = 1.175560
x_4 = 1.175560
After four iterations, the estimate converges to x ≈ 1.175560.
However, this solution cannot be expressed in terms of simple algebraic functions and must be obtained numerically.
However, this solution cannot be expressed in terms of simple algebraic functions and must be obtained numerically.
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Fermat's Last Theorem states that for any integer n greater than 2, there are no three integers a, b, and c that satisfy the equation a^n + b^n = c^n.
However, the equation given in this question, 4^x + 3^x = 5^x, is not in the form of Fermat's Last Theorem. Therefore, we cannot use Fermat's Last Theorem to solve this equation.
Instead, we can try to use algebraic techniques to solve for x. One possible approach is as follows:
We can rewrite the equation as:
(4/5)^x + (3/5)^x = 1
Letting y = (4/5)^x, we can rewrite the equation as:
y + (3/5*y) = 1
Solving for y, we get:
y = 5/8
Substituting back, we get:
(4/5)^x = 5/8
Taking the logarithm of both sides, we get:
x*log(4/5) = log(5/8)
Solving for x, we get:
x = log(5/8) / log(4/5)
Using a calculator or logarithmic tables, we get:
x ≈ 0.756
Therefore, the value of x is approximately 0.756.
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The final way to solve:
4^x+3^x=5^x is,
If (1+3)^x+(1+2)^x=(1+4)^x, then the value of
X=?........
Use binomial method ....
We get answer 2,✅
If (1+3)^x+(1+2)^x=(1+4)^x, then the value of
X=?........
Use binomial method
We can use the binomial theorem to expand (1+3)^x and (1+2)^x as follows:
(1+3)^x = 1^x + x31^(x-1) + (x choose 2)3^21^(x-2) + ... + (x choose k)3^k1^(x-k) + ... + 3^x
(1+2)^x = 1^x + x21^(x-1) + (x choose 2)2^21^(x-2) + ... + (x choose k)2^k1^(x-k) + ... + 2^x
Notice that all terms in the expansions have a factor of 1^k, so we can rewrite the given equation as:
1 + x*3^(x-1) + (x choose 2)*3^(x-2) + ... + (x choose k)*3^(x-k) + ...
1 + x*2^(x-1) + (x choose 2)*2^(x-2) + ... + (x choose k)2^(x-k) + ...
= 1 + x4^(x-1) + (x choose 2)*4^(x-2) + ... + (x choose k)*4^(x-k) + ...
Simplifying, we get:
2 + x3^(x-1) + x2^(x-1) + (x choose 2)(3^(x-2) + 2^(x-2)) + ...
= x4^(x-1) + (x choose 2)*4^(x-2) + ...
Notice that all terms except the first have a factor of 4^k, so we can rewrite the equation as:
2 + x3^(x-1) + x2^(x-1) + (x choose 2)*(3^(x-2) + 2^(x-2))
= 4^x - (x choose 0) - (x choose 1)*4^(x-1)
Simplifying further, we get:
2 + x3^(x-1) + x2^(x-1) + (x choose 2)(3^(x-2) + 2^(x-2)) = 34^(x-1) - x*4^(x-1)
Adding x*4^(x-1) to both sides, we get:
2 + x3^(x-1) + x2^(x-1) + (x choose 2)(3^(x-2) + 2^(x-2)) + x4^(x-1) = 3*4^(x-1)
Dividing both sides by 4^(x-1), we get:
2/4^(x-1) + x*(3/4)^{x-1} + x*(2/4)^{x-1} + (x choose 2)*[(3/4)^{x-2} + (2/4)^{x-2}]
= 3
Now, let's focus on the left-hand side of the equation. Notice that all terms except the first have a factor of (3/4)^{k-1} or (2/4)^{k-1}. Therefore, we can rewrite the equation as:
2/4^(x-1) + (3/4)^{x-1}[x + (x choose 2)(3/4) + (x choose 3)(3/4)^2 + ...] + (2/4)^{x-1}[x + (x choose 2)(2/3) + (x choose 3)*(2/3)^2 + ...] = 3
Now, we can simplify the terms inside the square brackets using the binomial theorem:
(3/4)^{x-1}[x + (x choose 1)(3/4) + (x choose 2)(3/4)^2 + ...]
= (2/4)^{x-1}[x + (x choose 1)(2/3) + (x choose 2)(2/3)^2 + ...]
Notice that the terms inside the square brackets are the expansions of (3/4 + 2/4)^x and (3/4 + 1/4)^x, respectively. Therefore, we can simplify the equation further:
2/4^(x-1) + (3/4)^{x-1}(5/4)^x + (2/4)^{x-1}(4/3)^x = 3
Let's rewrite the equation in terms of the variable y = (4/3)^x:
2/4^(x-1) + (3/4)^{x-1}(5/4)^x + (2/4)^{x-1}y = 3
Multiplying both sides by 4^(x-1), we get:
2 + 3^x*(5/4)^x + 2^xy = 34^(x-1)
Substituting y = (4/3)^x, we get:
2 + 3^x*(5/4)^x + 2^x*(4/3)^x = 3*4^(x-1)
Multiplying both sides by 3^x, we get:
23^x + 5^x + 2^x(3/2)^x = 3^x*4^(x-1)
Dividing both sides by 2^x*3^x, we get:
(2/3)^x + (5/6)^x = 2
Let's define f(x) = (2/3)^x + (5/6)^x. Then, we need to find the value of x such that f(x) = 2.
Note that f(x) is a decreasing function for x > 0, since:
f'(x) = -ln(2)(2/3)^x - ln(5/6)(5/6)^x < 0 for x > 0
Therefore, f(x) = 2 has at most one solution. We can see that x = 2 is a solution, since:
f(2) = (2/3)^2 + (5/6)^2 = 4/9 + 25/36 = 2
Therefore, the value of x is x = 2.
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