求解方程$(5x+1)^2 = 0.2\sqrt{5^x}$，对数变换后遇阻求指导

Hey there! Let's break down this problem and fix the misstep in your logarithm work first—then we'll talk about how to find solutions since this is a type of equation that doesn't have a nice "algebraic" answer.

Fixing Your Logarithm Step #

Your initial rewrite to $(5x+1)^2 = \frac{1}{5} \cdot 5^\frac{x}{2}$ is correct, but when you took logarithms, you missed applying $\ln$ to the entire right-hand side. You can't just take $\ln$ of the $5^\frac{x}{2}$ part while ignoring the $\frac{1}{5}$—logarithms apply to the whole term. The correct expansion would be:

$$\ln\left((5x+1)^2\right) = \ln\left(\frac{1}{5} \cdot 5^\frac{x}{2}\right)$$
Using logarithm rules ($\ln(ab)=\ln a + \ln b$, $\ln(a^b)=b\ln a$, $\ln(1/a)=-\ln a$), this becomes:
$$2\ln|5x+1| = -\ln 5 + \frac{x}{2}\ln 5$$

Why You Got Stuck #

The problem here is that the left side has $\ln|5x+1|$, a logarithmic function of $x$, and the right side is a linear function of $x$. This kind of equation is called a transcendental equation—it mixes polynomial, exponential, and logarithmic terms in a way that can't be solved using basic algebra or elementary functions. There's no way to rearrange this into a clean $x = \text{something}$ with just square roots, exponentials, etc.

Finding Numerical Solutions #

Since we can't get an exact elementary solution, we use numerical methods to approximate the root(s). Let's start by defining the function:
$$f(x) = (5x+1)^2 - 0.2\sqrt{5^x}$$
We need to find $x$ where $f(x)=0$.

First, narrow down where roots might be by testing values:

• $x=-0.1$: $f(-0.1) = (0.5)^2 - 0.2\sqrt{5^{-0.1}} ≈ 0.25 - 0.1866 = 0.0634 > 0$
• $x=-0.15$: $f(-0.15) = (0.25)^2 - 0.2\sqrt{5^{-0.15}} ≈ 0.0625 - 0.1782 = -0.1157 < 0$

So there's a root between $x=-0.15$ and $x=-0.1$. We can use the Newton-Raphson method to refine this approximation quickly:

1. Compute the derivative of $f(x)$:
   $$f'(x) = 10(5x+1) - 0.1\ln 5 \cdot 5^\frac{x}{2}$$
2. Start with an initial guess $x_0=-0.1$:
   • $f(x_0)≈0.0634$, $f'(x_0)≈5 - 0.1503=4.8497$
   • Next iteration: $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} ≈ -0.1 - \frac{0.0634}{4.8497}≈-0.1131$
3. Repeat with $x_1=-0.1131$:
   • $f(x_1)≈0.0054$, $f'(x_1)≈4.345 - 0.1476=4.1974$
   • $x_2 = -0.1131 - \frac{0.0054}{4.1974}≈-0.1144$
4. Checking $x=-0.1144$:
   • Left side: $(5(-0.1144)+1)^2=(0.428)2≈0.1832$
   • Right side: $0.2\sqrt{5^{-0.1144}}≈0.2(0.916)≈0.1832$

That's a solid approximation—$x≈-0.114$ is our real solution.

Checking for Other Roots #

Testing larger $x$ values (like $x=0,1,10$) shows $f(x)$ stays positive, and for $x < -0.2$, $f(x)$ is also positive. So this is the only real root of the equation.

内容的提问来源于stack exchange，提问作者leong seng cheong