Question

Use properties of exponents to find the value of log8 1 .(1 point

Answer 1

The value of x in log2x=1 is which of the following?(1 point)

Answer 2

If 10^logx=1/1032 , what is the value of x?(1 point)

Answer 3

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 \(\boxed{0}\).

Answer 4

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 \( \boxed{2} \).

Answer 5

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

\[
\boxed{10^{-32}}.
\]