If log a(ab) = x, then what is log b(ab)
The question asks us to find the value of $\log_b(ab)$ given that $\log_a(ab) = x$. This problem involves manipulating logarithmic expressions and changing the base of a logarithm.
We are given:
$\log_a(ab) = x$
We need to find the value of:
$\log_b(ab)$
To solve this problem, we will use the following fundamental properties of logarithms:
Let's start with the given equation: $\log_a(ab) = x$.
Step 1: Apply the Product Rule to the given equation.
Using the product rule, $\log_a(ab)$ can be written as $\log_a(a) + \log_a(b)$.
So, the given equation becomes:
$\log_a(a) + \log_a(b) = x$
Step 2: Simplify using the Logarithm of Base property.
We know that $\log_a(a) = 1$. Substituting this into the equation:
$1 + \log_a(b) = x$
Step 3: Isolate $\log_a(b)$.
Subtracting 1 from both sides, we get:
$\log_a(b) = x - 1$
This gives us a relationship between the bases a and b.
Step 4: Now consider the expression we need to find: $\log_b(ab)$.
Apply the Product Rule to $\log_b(ab)$:
$\log_b(ab) = \log_b(a) + \log_b(b)$
Step 5: Simplify using the Logarithm of Base property.
We know that $\log_b(b) = 1$. Substituting this into the expression:
$\log_b(ab) = \log_b(a) + 1$
Step 6: Relate $\log_b(a)$ to $\log_a(b)$.
Using the change of base formula, we know that $\log_b(a) = \frac{1}{\log_a(b)}$.
Step 7: Substitute the value of $\log_a(b)$ from Step 3 into the expression for $\log_b(a)$.
We found $\log_a(b) = x - 1$. So,
$\log_b(a) = \frac{1}{x - 1}$
Step 8: Substitute the value of $\log_b(a)$ from Step 7 into the expression from Step 5.
We have $\log_b(ab) = \log_b(a) + 1$. Substituting $\log_b(a) = \frac{1}{x-1}$:
$\log_b(ab) = \frac{1}{x - 1} + 1$
Step 9: Simplify the expression.
To simplify, find a common denominator, which is $(x-1)$.
$\frac{1}{x - 1} + 1 = \frac{1}{x - 1} + \frac{x - 1}{x - 1}$
Combine the numerators over the common denominator:
$\frac{1 + (x - 1)}{x - 1} = \frac{1 + x - 1}{x - 1} = \frac{x}{x - 1}$
Thus, $\log_b(ab) = \frac{x}{x - 1}$.
Based on our calculations, if $\log_a(ab) = x$, then $\log_b(ab) = \frac{x}{x - 1}$.
| Given Information | To Find | Result |
|---|---|---|
| $\log_a(ab) = x$ | $\log_b(ab)$ | $\frac{x}{x - 1}$ |
| Property Name | Formula | Usage in this Problem |
|---|---|---|
| Product Rule | $\log_c(MN) = \log_c(M) + \log_c(N)$ | Used to expand $\log_a(ab)$ and $\log_b(ab)$ |
| Logarithm of Base | $\log_c(c) = 1$ | Used to simplify $\log_a(a)$ and $\log_b(b)$ |
| Change of Base (Reciprocal) | $\log_c(d) = \frac{1}{\log_d(c)}$ | Used to relate $\log_b(a)$ to $\log_a(b)$ |
The base of a logarithm tells us which number is being raised to a power. For example, $\log_{10}(100)$ asks "10 to what power equals 100?". The answer is 2, so $\log_{10}(100) = 2$. Similarly, $\log_2(8)$ asks "2 to what power equals 8?". The answer is 3, so $\log_2(8) = 3$.
The change of base formula is very useful because it allows us to convert a logarithm with an inconvenient base to one with a more convenient base (like base 10 or base e, which are available on calculators). The formula $\log_c(d) = \frac{\log_k(d)}{\log_k(c)}$ shows that the ratio of logs with a common base 'k' gives the log in base 'c'. A special case is when we swap the base and the argument, $\log_c(d) = \frac{1}{\log_d(c)}$, which is what we used to relate $\log_a(b)$ and $\log_b(a)$.
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