Question

8*7^(2*x^2-x)-7*8^(2*x^2-x)=0

Answer 1

$8 \times {7}^{2 {x}^{2} - x} = 7 \times {8}^{2 {x}^{2} - x } \\( { \frac{7}{8} })^{2 {x}^{2} - x} = \frac{7}{8}$
2х^2-х=1
2х^2-х-1=0
х1=1
х2=-1/2

Answer 2

To solve the equation 8 * 7^(2x^2-x) - 7 * 8^(2x^2-x) = 0, we can simplify it step by step:

Step 1: Rewrite the equation using exponentiation properties.

8 * (7^(2x^2) * 7^(-x)) - 7 * (8^(2x^2) * 8^(-x)) = 0

Step 2: Use the fact that a^(m+n) = a^m * a^n and simplify the equation further.

8 * (49^x^(2x)) * (1/7^x) - 7 * (64^x^(2x)) * (1/8^x) = 0

Step 3: Rewrite 49 as 7^2 and 64 as 8^2.

8 * (7^(2x)^(2x)) * (1/7^x) - 7 * (8^(2x)^(2x)) * (1/8^x) = 0

Step 4: Apply the power of a power rule by multiplying the exponents.

8 * (7^(4x^2)) * (1/7^x) - 7 * (8^(4x^2)) * (1/8^x) = 0

Step 5: Combine the terms with the same base and simplify.

8 * 7^(4x^2 - x) - 7 * 8^(4x^2 - x) = 0

Now the equation is simplified to 8 * 7^(4x^2 - x) - 7 * 8^(4x^2 - x) = 0.

At this point, it is not possible to find an exact solution algebraically. However, you can solve it numerically using numerical methods or approximation techniques.