Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When the curves $y=\log_{10}x$ and $y=x-1$ are drawn in the $xy$ plane, how many times do they intersect?

To find intersection points eq.1 = eq. 2 $$\begin{align*} \log_{10}x &= x-1\\ 10^{x - 1} &= x \tag{a} \end{align*}$$

Answer would be no. of solutions (a) has. One of them i.e. 1 is easy to make out. You could check the degree of equation, that too is unclear (atleast to me).

Any suggestions on how to solve this (apart from plotting and checking)?

share|cite|improve this question
up vote 2 down vote accepted

Solving the equation $10^{x-1} = x$ exactly is difficult.

But you don't have to solve it exactly in order to figure out how many times the two graphs meet.

First, note that $y=\log_{10}x$ is only defined on the positive real numbers. So we can restrict ourselves to $(0,\infty)$.

Then, consider the function $f(x) = x-1-\log_{10}x$. The derivative of the function is $$f'(x) = 1 - \frac{1}{\ln(10)x}.$$ The derivative is positive if $x\gt \frac{1}{\ln(10)}$, and negative if $x\lt \frac{1}{\ln(10)}$. That means that the function $f(x)$ is decreasing on $(0,\frac{1}{\ln(10)})$, and is increasing on $(\frac{1}{\ln 10},\infty)$.

As $x\to 0^+$, we have $f(x)\to\infty$ (since $\log_{10}(x)\to-\infty$). At $x=\frac{1}{\ln(10)}$, we have $f(x)\approx -0.2035$; and as $x\to\infty$, $f(x)\to\infty$. So the function crosses the $x$-axis somewhere between $0$ and $\frac{1}{\ln(10)}\approx 0.4343$, and then again somwhere after $\frac{1}{\ln(10)}$ (well, at $x=1$, to be precise). And that's it.

So there are exactly two intersections.

share|cite|improve this answer
Neat Answer for it. – S. Snape Jun 19 '12 at 6:20

As you say, the obvious solution is at $x=1$. There is one other solution between $0$ and $1$ which is not easy to describe, but you can see its existence easily by noting that $x$ has constant derivative while $10^{x-1}$ has very small derivative for small $x$, thus they must intersect somewhere below $1$ (there derivatives are equal at $1$ and $10^{x-1}$ is always growing faster than $x$ above $1$). These are the only two solutions as the derivative of $10^{x-1}-x$ is $\ln 10\times 10^{x-1}-1$ which is monotonically increasing thus has only $1$ zero, and by Rolle's theorem if there were more than $2$ intersections there would be more than one zero.

share|cite|improve this answer

We have $\log_{10}x=\frac{\ln x}{\ln 10}$. Rewrite our equation as $$\ln x=(\ln 10)(x-1).$$ Note that $\ln 10\approx 2.3$. We look more generally at the equation $$\ln x=a(x-1),$$ where $a \gt 1$.

Let $f(x)=a(x-1)-\ln x$. We use standard tools from the calculus to analyze the behaviour of the curve $y=f(x)$.

We have $f'(x)=a-\frac{1}{x}$. Thus $f'(x)$ is $0$ at $x=1/a$, positive for $x \gt 1/a$, and negative for $0\lt x\lt 1/a$. It follows that $f$ is decreasing in the interval $(0,1/a)$, and then increasing. So $f$ attains a local minimum at $x=1/a$.

Since $f(1)=0$, and $1 \gt 1/a$, $f$ is positive for $x \gt 1$. It is easy to see that $f$ is positive when $x$ is close to $0$. The function $f$ reaches a minimum at $x=1/a$. The value of $f(a)$ must be negative, since $f$ is steadily increasing after $x=1/a$, but reaches $0$ at $x=1$. So $f(x)=0$ for some $x$ between $0$ and $1/a$.

It follows that the original equation has two roots, one between $0$ and $1/a$, and the other at $1$.

share|cite|improve this answer

Hint : Rate of growth of $\log x > x - 1$ for $x < 1$. Similarly, it's lesser for $x > 1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.