You could solve this by using your calculator. Remember that you will have to translate this into:

\(\displaystyle \frac{log(64)}{log(2)}\)

Another way you can solve it is by noticing that \(\displaystyle 64=2^6\)

This means you can rewrite your logarithm:

\(\displaystyle log_2(2^6)\)

Applying logarithm rules, you can factor out the power:

\(\displaystyle log_2(2^6)=6log_2(2)\)

For any value \(\displaystyle n\), \(\displaystyle log_n(n) = 1\). Therefore, \(\displaystyle log_2(2) = 1\). So, your answer is \(\displaystyle 6\).

To solve an exponential equation like this, you need to use logarithms.  This can be translated into:

\(\displaystyle log_5(60)=x+1\)

Now, remember that your calculator needs to have this translated.  The logarithm \(\displaystyle log_5(60)\) is equal to the following:

\(\displaystyle \frac{log(60)}{log(5)}\), which equals approximately \(\displaystyle 2.54\).

Remember that you have the equation:

\(\displaystyle 2.54 = x + 1\)

Thus, \(\displaystyle x=1.54\).

In order to solve a question like this, you will need to use logarithms. First, start by converting this into a basic logarithm:

\(\displaystyle log_3(44)=2x-4\)

Recall that you need to convert \(\displaystyle log_3(44)\) for your calculator:

\(\displaystyle \frac{log(44)}{log(3)}\), which equals approximately \(\displaystyle 3.44\)

Thus, you can solve for \(\displaystyle x\):

\(\displaystyle 3.44=2x-4\)

\(\displaystyle 2x = 7.44\)

\(\displaystyle x=3.72\)

The general function that defines this compounding interest is:

\(\displaystyle value = 1.045^t* 1400\), where \(\displaystyle t\) is the number of years.

What we are looking for is:

\(\displaystyle 1.045^t* 1400 = 5000\)

You can solve this using a logarithm.  First, isolate the variable term by dividing both sides:

\(\displaystyle 1.045^t =\frac{5000}{1400}\)

Which is:

\(\displaystyle 1.045^t =\frac{25}{7}\)

Next, recall that this is the logarithm:

\(\displaystyle t = log_1_._0_4_5(\frac{25}{7})\)

For this, you will need to do a base conversion:

\(\displaystyle log_1_._0_4_5(\frac{25}{7}) = \frac{log(\frac{25}{7})}{log(1.045)}\)

This is \(\displaystyle 28.9199397858296...\)

This means that it will take \(\displaystyle 29\) years.  \(\displaystyle 28\) is too few and at the end of \(\displaystyle 29\), you will have over \(\displaystyle 5000\).

Remember that you will need to calculate your logarithm by doing a base conversion. This is done by changing \(\displaystyle log_4_._5(381)\) into:

\(\displaystyle \frac{log(381)}{log(4.5)}\)

Using your calculator, you can find this to be:

\(\displaystyle 3.95112604435386...\) or approximately \(\displaystyle 3.95\)

The first step of this problem is to find

\(\displaystyle \log_4(16)\) by expanding to the formula

\(\displaystyle 4^y = 16.\)  

y is found to be 2.  The next step is to plug y in to the second log.  

\(\displaystyle log_2(2)\), which expands to

\(\displaystyle 2^x = 2,\)

\(\displaystyle x = 1\)

\(\displaystyle log_3(81)\) 

expands to

\(\displaystyle 3^y = 81, y = 4\)

\(\displaystyle log_2(4)\) 

expands to

\(\displaystyle 2^x = 4, x = 2\)

Here, we need to make use of some logarithm identities:  \(\displaystyle \textup{log}(\textup{a}^\textup{n})=\textup{n}*\textup{log(a)}log(a)+\textup{log}(\textup{b})=\textup{log(ab)}\)  

\(\displaystyle \textup{and } \textup{log(a)}-\textup{log(b)}=log(\frac{a}{b}).\)

Therefore, putting all of those things together, we get the final answer of \(\displaystyle \textup{log a}^2+\textup{log b}+\textup{log c}^3= \textup{log}(\frac{2\textup{ab}}{3\textup{c}}).\)

This is a test of translating logarithmic/exponential properties, with the key here being to realize that

\(\displaystyle \log_{4}x = \frac{5}{2}\) is equivalent to \(\displaystyle x = 4^{5/2}\).

With that in mind, here is how it works out:

\(\displaystyle 4^{5/2} = (\sqrt[2]{4})^{5} = 2^{5} = 2*2*2*2*2 = 32\)

Hence, \(\displaystyle x = 32\).

A is true in two ways. You can use the fact that if a logarithm has no base, you can use base 10, or you can use the fact that you can use this property:

\(\displaystyle \log_{a}b = \frac{\log_{c}b}{\log_{c}a}\)

B is a simple change of base application, and C is simply computing the logarithm.

\(\displaystyle \log_{5}25 = \frac{\log_{10}25}{\log_{10}5}\)

\(\displaystyle \log_5 25\rightarrow \log_ab=c \rightarrow a^c=b\)

\(\displaystyle \\ \log_5 25\\ 5^x=25\\5^2=25\\x=2\)