For what values of x is log subscript 0 point 7 5 baseline left parenthesis x plus 3 right parenthesis greater than log subscript 0 point 5 baseline left parenthesis x plus 3 right parenthesis?

(1 point)
Responses

negative 2 less than x less than infinity
Image with alt text: negative 2 less than x less than infinity

negative 3 less than x less than negative 2
Image with alt text: negative 3 less than x less than negative 2

4 less than x less than infinity
Image with alt text: 4 less than x less than infinity

3 less than x less than 4

1 answer

To solve the inequality \( \log_{0.75}(x + 3) > \log_{0.5}(x + 3) \), we start by recognizing that both logarithmic functions are defined only if \( x + 3 > 0 \), which means \( x > -3 \).

Next, we can rewrite the logarithm into a common base. In this case, we can convert the logarithm to base 10 or natural logarithm for easier comparison. However, notice that since \( 0.75 < 1 \) and \( 0.5 < 1 \), the logarithm will be a decreasing function:

  1. \( \log_{0.75}(x + 3) = \frac{\log(x + 3)}{\log(0.75)} \)
  2. \( \log_{0.5}(x + 3) = \frac{\log(x + 3)}{\log(0.5)} \)

Since both logarithmic functions are decreasing, we can find the point where they are equal:

Set: \[ \log_{0.75}(x + 3) = \log_{0.5}(x + 3) \]

Because both sides can be multiplied by the positive value \( \log(0.75) \) and \( \log(0.5) \) (logarithmic bases less than 1 lead to a flip in the inequality):

\[ \log(x + 3) \cdot \log(0.5) < \log(x + 3) \cdot \log(0.75) \]

This means whenever \( \log(x + 3) \) is positive, \( \log_{0.75}(x + 3) \) is greater than \( \log_{0.5}(x + 3) \); hence we need to find \( x + 3 > 1 \) or \( x > -2 \).

So the solution to the inequality is: \[ x > -2 \]

Now remembering the previous condition that \( x + 3 > 0 \) which translates to \( x > -3 \), we find that:

\( x > -2 \) is the more restrictive of the two conditions.

Therefore, the range of values for \( x \) is: \[ \text{negative 2 less than x less than infinity} \]

So the answer is: negative 2 less than x less than infinity.