I have two short questions which I think I know the answer-could you please check?
If I write log(b)x - 2log(b)y as a single logarithm it would be log(b) x/y^2, correct? If not please help
and Is the function f(x) = 1/x a exponential function -I think it would be because it could be a number to the negative power, correct or would that be false
The function f(x) = logx is a logarthmic function-I think this is false because it should be written as log(then a number)x, correct
Could one of you please check when you have time.
Thank you
correct, if you meant logb(x/y^2)
F(x)=x^-1 is an inverse function, not an exponential function. Exponential functions are functions which can be written as f(x)=k e^ax
for instance f(x)=3e^-5
or f(x)=12(1-e^(-4x) )
Can it be applied to g(x)=3 a^bx ?
many people include this form in the definition of exponential functions.
I would call f(x)=log x a log function, it is a log whether or not the base is defined. changing bases just changes the slopes of the log curves.
Thank you for the explanations- I appreciate it
I just noticed something-I had asked about if f(x) = 1/x is an exponential function and I think you rewrote it as f(x) = x^-1. Therefore, f(x) = 1/x is NOT an exponential function, correct?
correct, not an exponential function.
Thank you-could you look at my new post-I just needed clarification about f(x) = logx you said this would be true but doesn't it need parentheses or a separation between log and x for this to be correct?
Thank you
Let's check the answers to your questions one by one:
1. If you have the expression log(b)x - 2log(b)y and want to rewrite it as a single logarithm, you need to use the logarithmic properties. One of the properties states that when you subtract logarithms with the same base, it is equivalent to dividing the corresponding arguments. Therefore:
log(b)x - 2log(b)y = log(b)(x) - log(b)(y^2)
Now, another logarithmic property states that when you have the logarithm of a power, it is equivalent to multiplying the exponent by the logarithm of the base. In this case, we have y^2 as the argument of the second logarithm, so we can apply this property to rewrite it as:
log(b)(x) - log(b)(y^2) = log(b)(x) - 2log(b)(y) = log(b)(x/y^2)
So, your answer is correct.
2. The function f(x) = 1/x is not an exponential function. An exponential function is defined as a function where the variable appears in the exponent. In this case, the variable x appears in the denominator, not as an exponent. Therefore, f(x) = 1/x is not an exponential function.
Your initial understanding is correct. The function f(x) = 1/x can be considered a power function, specifically the function f(x) = x^(-1). Power functions are a broader category that includes exponential functions.
3. Your statement about the function f(x) = log(x) being false is correct. The function f(x) = log(x) is indeed a logarithmic function. In logarithmic functions, the variable, in this case, x, appears as the argument being logarithmically transformed. It is important to note that the base of the logarithm is typically assumed to be 10 unless otherwise specified, so f(x) = log(x) can be interpreted as log10(x).
In summary:
- log(b)x - 2log(b)y can be simplified to log(b)(x/y^2).
- The function f(x) = 1/x is not an exponential function but a power function.
- The function f(x) = log(x) is a logarithmic function.