Question
Use properties of exponents to find the value of log8 1 .(1 point
Answers
GPT-4o mini
To find the value of \(\log_8 1\), we can use the property of logarithms that states:
\[
\log_b a = c \quad \text{if and only if} \quad b^c = a
\]
In this case, we want to solve for \(\log_8 1\). So we set:
\[
\log_8 1 = c
\]
This implies:
\[
8^c = 1
\]
We know that any non-zero number (such as 8) raised to the power of 0 is 1:
\[
8^0 = 1
\]
This means that:
\[
c = 0
\]
Thus, we find:
\[
\log_8 1 = 0
\]
Therefore, the value of \(\log_8 1\) is \(\boxed{0}\).
\[
\log_b a = c \quad \text{if and only if} \quad b^c = a
\]
In this case, we want to solve for \(\log_8 1\). So we set:
\[
\log_8 1 = c
\]
This implies:
\[
8^c = 1
\]
We know that any non-zero number (such as 8) raised to the power of 0 is 1:
\[
8^0 = 1
\]
This means that:
\[
c = 0
\]
Thus, we find:
\[
\log_8 1 = 0
\]
Therefore, the value of \(\log_8 1\) is \(\boxed{0}\).
The value of x in log2x=1 is which of the following?(1 point)
GPT-4o mini
To solve the equation \( \log_2 x = 1 \), we can use the definition of logarithms. Specifically, the equation \( \log_b a = c \) is equivalent to \( b^c = a \).
In this case, we have:
\[
\log_2 x = 1
\]
This means:
\[
2^1 = x
\]
Calculating the power gives us:
\[
x = 2
\]
Therefore, the value of \( x \) is \( \boxed{2} \).
In this case, we have:
\[
\log_2 x = 1
\]
This means:
\[
2^1 = x
\]
Calculating the power gives us:
\[
x = 2
\]
Therefore, the value of \( x \) is \( \boxed{2} \).
If 10^logx=1/1032 , what is the value of x?(1 point)
GPT-4o mini
To solve the equation
\[
10^{\log x} = \frac{1}{10^{32}},
\]
we can simplify the right side:
\[
\frac{1}{10^{32}} = 10^{-32}.
\]
Now we have:
\[
10^{\log x} = 10^{-32}.
\]
Since the bases are the same, we can set the exponents equal to one another:
\[
\log x = -32.
\]
Next, we can rewrite this in exponential form:
\[
x = 10^{-32}.
\]
Thus, the value of \( x \) is
\[
\boxed{10^{-32}}.
\]
\[
10^{\log x} = \frac{1}{10^{32}},
\]
we can simplify the right side:
\[
\frac{1}{10^{32}} = 10^{-32}.
\]
Now we have:
\[
10^{\log x} = 10^{-32}.
\]
Since the bases are the same, we can set the exponents equal to one another:
\[
\log x = -32.
\]
Next, we can rewrite this in exponential form:
\[
x = 10^{-32}.
\]
Thus, the value of \( x \) is
\[
\boxed{10^{-32}}.
\]