7^x-5x+6=1 p.s:х-5х+6-это степень

To solve the equation $7^{x^2-5x+6}=1$, we need to apply the properties of exponents. 1. Identify the Core Property Any non-zero number raised to the power of $0$ is equal to $1$. Mathematically, this is expressed as: $a^0=1(\text{where}a\ne 0)$In your equation, the base is $7$. Therefore, for the expression to equal $1$, the exponent must be equal to $0$. 2. Set up the Quadratic Equation By equating the exponent to zero, we get: $x^2-5x+6=0$3. Solve the Quadratic Equation We can solve this using the quadratic formula or by factoring. Factoring Method: We look for two numbers that multiply to $\mathrm{+6}$ and add up to $-5$. These numbers are $-2$ and $-3$. $(x-2)(x-3)=0$Setting each factor to zero:

• $x-2=0{\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{2}$ $x-3=0{\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{3}$

4. Verification

• If $x=2$: $7^{2^2-5\left(2\right)+6}=7^{4-10+6}=7^0=1$ If $x=3$: $7^{3^2-5\left(3\right)+6}=7^{9-15+6}=7^0=1$

Final Answer: The solutions to the equation are $x=2$ and $x=3$. Would you like me to solve a similar equation involving a different base or a more complex exponent?