UME 1999 (type B), Question 3: If \( log_{8}10=x\), evaluate \(log_{8}5\) in terms of \(x\)

 Solution

We are given that:

$$\log_8 10 = x$$

We are asked to find $\log_8 5$ in terms of $x$. This means we need to express $\log_8 5$ using $x$, which we know is equal to $\log_8 10$.

Step 1: What is a Logarithm?

Before we proceed, let’s quickly recall what a logarithm is.

A logarithm is simply the opposite (inverse) of an exponent. For example, if:

$$a^b = c$$

Then:

$$\log_a c = b$$

This means that $\log_a c$ tells us what power we need to raise $a$ to in order to get $c$.

Step 2: Using the Given Information

We are told that:

$$\log_8 10 = x$$

This means that the logarithm of 10 with base 8 is equal to $x$. In other words, $8^x = 10$.

Now, our goal is to find $\log_8 5$ in terms of $x$. We want to express $\log_8 5$ using the fact that $\log_8 10 = x$.

Step 3: Change of Base Formula

One powerful tool we can use is called the change of base formula. It helps us change the base of the logarithm to a new base we’re more comfortable with (such as base 2).

The change of base formula is:

$$\log_b a = \frac{\log_c a}{\log_c$$

This means that to calculate $\log_b a$, we can divide $\log_c a$ by $\log_c b$, where $c$ is any base we choose (like base 2 or base 10).

Step 4: Applying the Change of Base Formula

We are given $\log_8 10 = x$, and we want to find $\log_8 5$. Let's use the change of base formula with base 2 (since base 2 is often easier to work with).

First, let’s rewrite $\log_8 10$ using base 2:

$$\log_8 10 = \frac{\log_2 10}{\log_2$$

This is because $8$ is the same as $2^3$, and we can work with base 2.

Now, from the problem, we know that:

$$\log_8 10 = x$$

So we can write:

$$x = \frac{\log_2 10}{\log$$

We also know that $\log_2 8 = 3$, because $8 = 2^3$. So the equation becomes:

$$x = \frac{\log_2 10}{3}$$Multiplying both sides by 3:$$\log_2 10 = 3x$$

Step 5: Express $\log_8 5$ in Terms of $x$

Now, we can find $\log_8 5$. Using the change of base formula again:

$$\log_8 5 = \frac{\log_2 5}{\log_$$

Since $\log_2 8 = 3$, we have:

$$\log_8 5 = \frac{\log_2 5}{3$$

​Step 6: Relating $\log_2 5$ to $\log_2 10$

We know that $10 = 2 \times 5$, so we can use the logarithm property that says:

$$\log_2 10 = \log_2 (2 \times 5)$$

This can be broken down as:

$$\log_2 10 = \log_2 2 + \log_2 5$$

Since $\log_2 2 = 1$, we get:

$$\log_2 10 = 1 + \log_2 5$$

Now, we know from Step 4 that $\log_2 10 = 3x$, so we can substitute this into the equation:

$$3x = 1 + \log_2 5$$

To solve for $\log_2 5$, subtract 1 from both sides:

$$\log_2 5 = 3x - 1$$

Step 7: Final Answer

Now, substitute ${\log }_{2}5=3x-1$ into the formula for ${\log }_{8}5$:

$$\log_8 5 = \frac{3x - 1}{3}$$Simplify:$$\log_8 5 = x - \frac{1}{3}$$So, we have successfully expressed $\log_8 5$ in terms of $x$ as:$$\log_8 5 = x - \frac{1}{3}$$ 

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