Question

5^(2√х) + 125 = 6 ∙5^(√х+1)

Answer 1

Ответ:

${5}^{2 \sqrt{x} } = {({5}^{ \sqrt{x} } ) }^{2}$

${5}^{ \sqrt{x} + 1 } = {5}^{ \sqrt{x} } \times 5$

И мы заменяем 5^√х на t и решаем квадратное уравнение, а потом делаем обратную замену)

Answer 2

To solve the equation $5^{2\sqrt{x}} + 125 = 6 \cdot 5^{\sqrt{x} + 1}$ for $x$, follow these steps:

Step 1: Rewrite the equation to isolate one of the terms involving $5^{\sqrt{x}}$: $5^{2\sqrt{x}} = 6 \cdot 5^{\sqrt{x} + 1} - 125$

Step 2: Use the properties of exponents to manipulate the equation further. Recall that $a^{m+n} = a^m \cdot a^n$ and $a^{mn} = (a^m)^n$: $5^{2\sqrt{x}} = 6 \cdot 5^{\sqrt{x}} \cdot 5^1 - 125$ $5^{2\sqrt{x}} = 30 \cdot 5^{\sqrt{x}} - 125$

Step 3: Now, let's make a substitution to simplify the equation. Let $y = 5^{\sqrt{x}}$, so the equation becomes: $y^2 = 30y - 125$

Step 4: Rearrange the equation into a quadratic equation form: $y^2 - 30y + 125 = 0$

Step 5: Solve the quadratic equation for $y$. You can factor the quadratic or use the quadratic formula: $(y - 25)(y - 5) = 0$

This gives us two possible values for $y$:

1. $y = 25$
2. $y = 5$

Step 6: Substitute the values of $y$ back into the equation $y = 5^{\sqrt{x}}$ and solve for $x$: For $y = 25$: $25 = 5^{\sqrt{x}}$ Take the logarithm base 5 of both sides: $\log_5 25 = \sqrt{x}$ $2 = \sqrt{x}$ $x = 4$

For $y = 5$: $5 = 5^{\sqrt{x}}$ Take the logarithm base 5 of both sides: $1 = \sqrt{x}$ $x = 1$

So, the possible solutions for $x$ are $x = 1$ and $x = 4$. You should plug these solutions back into the original equation to verify their correctness.