log 5 �ã5
log7 (1/49)
Here are some examples to help you.
log2 32 = ?
2^? = 32
? = 5
log3 1/9 = ?
3^? = 1/9
? = -2
As you can see, if y = logb x, then b^y = x.
To solve the first problem, log 5 (‑5), we need to understand the properties of logarithms.
The logarithm is the inverse function of exponentiation. In other words, if we have the equation x = a^b, then the logarithmic form of this equation is log base a (x) = b.
In this case, we have log 5 (‑5). This means we need to find the power to which 5 must be raised in order to get ‑5. However, it's important to note that logarithms of negative numbers are undefined in the real number system.
So, log 5 (‑5) is undefined. There is no solution to this logarithmic equation in the real number system.
For the second problem, log7 (1/49), we can use the property of logarithms which states that log base a (1/x) = ‑log base a (x).
Applying this property to our problem, we can rewrite log7 (1/49) as ‑log7 (49). Now we need to find the value of ‑log7 (49).
To evaluate this expression, we need to understand that ‑log7 (49) means we are looking for the power to which 7 must be raised to get 49. In other words, we're finding the exponent to which 7 must be raised to get 49.
Since 49 is equal to 7^2, we know that ‑log7 (49) is equal to ‑2. Therefore, the solution to log7 (1/49) is ‑2.