# How many times do these curves intersect?

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)?

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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.

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Neat Answer for it. – Babak S. 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.

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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 $1$.

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

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