Q.Prove that the logarithmic function is increasing on $(0, \infty)$.

The key idea is that the derivative of $\log_a x$ is positive for $a > 1$, which directly implies monotonic increase.

Step 1: Consider the logarithmic function $f(x) = \log_a x$ with base $a > 1$, defined for $x > 0$.

Step 2: Differentiate: $f'(x) = \frac{1}{x \ln a}$.

Step 3: Since $x > 0$ and $\ln a > 0$ (because $a > 1$), the derivative $f'(x) > 0$ for all $x \in (0, \infty)$. …

Why This Works: The Concept of Monotonicity

A function is increasing on an interval if, whenever you take two points $x_1 < x_2$, the function values satisfy $f(x_1) < f(x_2)$. For the logarithm $f(x) = \log_a x$ (with base $a > 1$), we need to show that a larger input always gives a larger output.

The cleanest way to prove this is to use the fact that the logarithm is the inverse of the exponential function $g(x) = a^x$. And we already know that $a^x$ is strictly increasing when $a > 1$ — if you raise a bigger exponent, you get a bigger result. Since inverses of strictly increasing functions are also strictly increasing, the logarithm inherits this property.

But let's make this rigorous without assuming the inverse property.

Step-by-Step Proof

1. Set up what we need to prove.

Take any two positive numbers $x_1$ and $x_2$ such that $0 < x_1 < x_2$. We must show:

2. Use the definition of the logarithm.

Let $y_1 = \log_a x_1$ and $y_2 = \log_a x_2$. By definition, this means:

3. Translate the inequality into exponents.

We know $x_1 < x_2$, so:

4. Apply the monotonicity of the exponential.

The function $h(t) = a^t$ (with $a > 1$) is strictly increasing. This means:

5. Translate back to logarithms.

Since $y_1 = \log_a x_1$ and $y_2 = \log_a x_2$, the inequality $y_1 < y_2$ becomes:

6. Conclude. …

Method: Proving Strict Monotonicity via the Sign of the Derivative

This method applies whenever you must prove (not just locate) that a function is increasing or decreasing on a stated interval, especially when the function is built from a known family such as $\log_a x$ or $a^x$ and a condition on a parameter (like the base $a$) is what makes the proof work.

Steps

Step 1: Write down the domain and the derivative test you will use.

Note any domain restriction (here $x>0$, since $\log_a x$ needs a positive argument), and recall: if $f'(x) > 0$ for every $x$ in the interval, $f$ is strictly increasing there.

Step 2: Differentiate using the standard rule for the function's family, keeping any parameter symbolic.

Do not substitute a specific base — keep $a$ as a symbol so the proof covers the general statement asked for.

Step 3: Determine the sign of every factor in the derivative using the restrictions given. …

Common Mistakes

Mistake 1: Forgetting the base restriction $a > 1$.

Why it's wrong: the sign of $\ln a$ flips for $0 < a < 1$, which would make $f'(x) < 0$ and the function decreasing instead. Correct approach: always state explicitly that the proof depends on $a>1$; without it, the increasing conclusion doesn't hold.

Mistake 2: Differentiating $\log_a x$ as if it were $\ln x$.

Why it's wrong: dropping the $\ln a$ factor gives $f'(x) = 1/x$, which is only valid for the natural logarithm ($a=e$), not a general base. Correct approach: use the change-of-base derivative $\dfrac{1}{x\ln a}$ and only simplify to $1/x$ when $a=e$ is explicitly given. …