High School

Third law - Logarithm powers. Simplify the following:

(i) log_2 8^3
(ii) 4log_9 3
(iii) Log_5 1/125

Answer :

Let's simplify each of the logarithmic expressions step by step using logarithm powers and properties.

(i) [tex]\log_2 8^3[/tex]

Step 1: Use the power rule of logarithms, which states [tex]\log_b a^n = n \cdot \log_b a[/tex]. This means you can bring the exponent down as a coefficient.

[tex]\log_2 8^3 = 3 \cdot \log_2 8[/tex]

Step 2: Simplify [tex]\log_2 8[/tex]. Since 8 is a power of 2 (specifically, [tex]2^3[/tex]), we can rewrite it as:

[tex]\log_2 8 = \log_2 (2^3) = 3[/tex]

Step 3: Substitute back:

[tex]3 \cdot \log_2 8 = 3 \cdot 3 = 9[/tex]

Thus, [tex]\log_2 8^3 = 9[/tex].

(ii) [tex]4 \cdot \log_9 3[/tex]

In some problems, there might be specific properties or numbers involved to simplify, but in this case, there is no such simplification possible directly since no powers relate directly.

Thus, [tex]4 \cdot \log_9 3[/tex] stays as is unless more context or restrictions are applied.

(iii) [tex]\log_5 \frac{1}{125}[/tex]

Step 1: Use the property of logarithms that states [tex]\log_b \frac{1}{a} = -\log_b a[/tex]. First, rewrite the expression:

[tex]\log_5 \frac{1}{125} = -\log_5(125)[/tex]

Step 2: Simplify [tex]\log_5 125[/tex]. Since 125 is a power of 5 (specifically, [tex]5^3[/tex]), rewrite it using that power:

[tex]\log_5(125) = \log_5(5^3) = 3[/tex]

Step 3: Substitute back:

[tex]-\log_5 125 = -3[/tex]

Thus, [tex]\log_5 \frac{1}{125} = -3[/tex].

So, the final simplified forms are:

  1. [tex]\log_2 8^3 = 9[/tex]
  2. [tex]4 \cdot \log_9 3[/tex] (remains unchanged without further context)
  3. [tex]\log_5 \frac{1}{125} = -3[/tex]

These calculations use fundamental properties of logarithms to simplify each expression.