How to solve 5x=0.01^x

Question

I just want to know how to solve:

$$\ 5x=0.01^x$$

I have tried to use logarithms. It would be a huge help if someone could help because no matter what I do $\ x$ always gets stuck in a logarithm.

Edit: Sorry, I changed the questions to match.

Answer 1

In general, here are the steps $$\begin{align} ax &= b^x \\ \frac{ax}{b^x} &= 1 \\ axb^{-x} &= 1 \\ axe^{\ln b^{-x}} &= 1 \\ axe^{-x\ln b} &= 1 \\ xe^{-x\ln b} &= \frac{1}{a} \\ x(-\ln b)e^{-x\ln b} &= -\frac{\ln b}{a} \\ (-x\ln b)e^{(-x\ln b)} &= -\frac{\ln b}{a} \end{align}$$

Which implies that $$ W\left(-\frac{\ln b}{a}\right) = -x\ln b $$ Therefore $$ x = -\frac{W\left(-\dfrac{\ln b}{a}\right)}{\ln b} $$ Where $W(z)$ is the Lambert W Function.

Answer 2

if you take the logarithm, you get $\ln 5 + \ln x = x \ln (1/100) = -2x \ln10$

this has an approximate root at $ x = 0.116$ according to my ti-83 solver.

Answer 3

Note: Due to a typo in the original question the question I answered here is slightly different than the one asked by the OP. However the method described will work to answer the question.

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As pointed out by others there isn't a solution involving the elementary functions you are familiar with. If you are a pre-calculus student then you are likely expected to solve the equation by finding the intersection of the curves on your graphing calculator.

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That said there is a recursive approach for producing approximate answers to this equation which isn't too hard to understand. We will start by graphing both sides of the equation. The plot below was produced by Wolfram Alpha.

We can see that the intersection of the curves occurs somewhere between $x=0$ and $x=0.5$. We are going to start by guessing an answer which we know is wrong, but not too far from the actual answer. In this case we could start with $x_1=0.3$.

Now notice that $5x_1$ is larger than $.1^{x_1}$ if $x_1$ is larger than the intersection. Also notice that $5x_1$ is smaller than $.1^{x_1}$ if $x_1$ is smaller than the intersection. This motivates us to consider the difference,

$$ \Delta(x)= .1^x - 5x,$$

which is positive is $x$ is too big and negative is $x$ is too small. We can correct our guess, $x_1$, by adding to it a fraction of $\Delta(x_1)$. This will give us a new guess, $x_2$, which is slightly better.

$$ x_2 = x_1 + .1 \Delta(x_1) = 0.3 +0.1(0.1^{0.3}-5 \cdot 0.3) \approx 0.2001 $$

Looking at the graphs again we can see that $x_2$ is much closer to the intersection. We can repeat this process to get better estimates.

$$ x_3 = x_2 + .1 \Delta(x_2) = 0.2001 +0.1(0.1^{0.2001}-5 \cdot 0.2001) \approx 0.163131 $$

$$ x_4 = x_3 + .1 \Delta(x_3) = 0.163131 +0.1(0.1^{0.163131}-5 \cdot 0.163131) \approx 0.150252 $$

$$ x_5 = x_4 + .1 \Delta(x_4) = 0.150252 +0.1(0.1^{0.150252}-5 \cdot 0.150252) \approx 0.145880 $$

$$ x_6 = x_5 + .1 \Delta(x_5) = 0.145880+0.1(0.1^{0.145880}-5 \cdot 0.145880) \approx 0.144490 $$

$$ x_7 = x_6 + .1 \Delta(x_6) = 0.144490+0.1(0.1^{0.144490}-5 \cdot 0.144490) \approx 0.143943 $$

$$ x_8 = x_7 + .1 \Delta(x_7) = 0.143943 +0.1(0.1^{0.0.143943 }-5 \cdot 0.143943 ) \approx 0.143760$$

$$ x_9 = x_8 + .1 \Delta(x_8) =0.143760 +0.1(0.1^{0.143760 }-5 \cdot 0.143760 ) \approx 0.143699$$

and so on.

The idea is to repeat the process until the digits in your calculator stop changing. If you know how to use excel then you can set this process up easily in a spread sheet and quickly produces very good estimates of the intersection point.