Animated Solution for Mathematics - Basic Mathematics: If ${\log }_{0.3}(x-1)<{\log }_{0.09}(x-1)$, then $x$ lies in the interval

Our First Boundary

Welcome, aspiring mathematician! Today, we are embarking on a journey to conquer a logarithmic inequality. Before we even touch the algebra, we must respect the golden rule of logarithms: the argument must be strictly positive.

The function ${\log }_{0.3}(x-1)$ is only defined if $x-1>0$, which means $x>1$. This is our 'playground.' Any solution we find must exist within this region.

If we ignore this, we are essentially trying to build a house on quicksand. So, let's mark $x=1$ as our vertical boundary and keep it in mind.

Seeing the Hidden Pattern

Now, look at the inequality:

The bases are $0.3$ and $0.09$. Do you see the connection? It is a beautiful piece of pattern recognition: $0.09=(0.3{)}^2$.

This is not a coincidence; it is a deliberate invitation to use the power property of logarithms. We know that ${\log }_{a^n}b=\frac{1}{n}{\log }_{a}b$. By applying this, we can transform the right-hand side into something much more manageable.

Simplifying the Expression

Let's apply that property. The right side becomes:

Now, our inequality looks like this:

Let's treat the log term as a single variable, say $Y$. We have $Y<\frac{1}{2}Y$. Subtracting $\frac{1}{2}Y$ from both sides, we get $\frac{1}{2}Y<0$, which simplifies to:

We have successfully reduced a complex-looking inequality into a simple, elegant form.

The Moment of Truth

This is where many students stumble, but you won't. We need to remove the logarithm to solve for $x$. We convert to exponential form: $x-1$ compared to $(0.3{)}^0$.

But wait! The base is $0.3$. Since $0<0.3<1$, the logarithmic function is strictly decreasing. This means that as the input increases, the output decreases.

To maintain the inequality, we MUST flip the sign. The less-than sign becomes a greater-than sign. So:

Since any non-zero number to the power of $0$ is $1$, we get $x-1>1$.

Bringing it All Together

Solving $x-1>1$ gives us $x>2$. But remember our domain from the beginning? We needed $x>1$.

Our final solution must satisfy both conditions: $x>1$ AND $x>2$. The intersection of these two sets is simply $x>2$.

Thus, the final interval is $(2,\infty )$. You have just navigated the traps, applied the properties, and arrived at the truth. Keep this rigor in your toolkit, and no inequality will ever stand in your way again!