Prove \(f(x)=\log_a x\) is a Bijection

📺 Video Explanation

📝 Question

Show that:

\[ f:\mathbb{R}_0^+\to\mathbb{R},\quad f(x)=\log_a x \]

where:

\[ a>0,\ a\neq1 \]

is a bijection.

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✅ Solution

🔹 Step 1: Prove One-One (Injective)

Assume:

\[ f(x_1)=f(x_2) \]

Then:

\[ \log_a x_1=\log_a x_2 \]

Therefore:

\[ x_1=x_2 \]

✔ Hence, one-one.

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🔹 Step 2: Prove Onto (Surjective)

Let:

\[ y\in\mathbb{R} \]

Need:

\[ \log_a x=y \]

This gives:

\[ x=a^y \]

Since:

\[ a^y>0 \]

So:

\[ x\in\mathbb{R}_0^+ \]

Thus every real number has a pre-image.

✔ Hence, onto.

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🎯 Final Answer

\[ \boxed{f(x)=\log_a x\text{ is bijective}} \]

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🚀 Exam Shortcut

• Log function is inverse of exponential
• Equal logs imply equal numbers
• Use \(x=a^y\) for onto proof