For any equation \(\displaystyle \log_{10}(y) = x\), \(\displaystyle 10^{x } = y\). Thus, we are trying to determine what power of 10 is 1000. \(\displaystyle 1000 = 10^3\), so our answer is 3. 

Take the common logarithm of both sides, and take advantage of the property of the logarithm of a power:

\(\displaystyle 7 ^{x} = 1,000\)

\(\displaystyle 7 ^{x} = 10^{3}\)

\(\displaystyle \log 7 ^{x} = \log 10^{3}\)

\(\displaystyle x \log 7 = 3 \log 10\)

\(\displaystyle x \log 7 = 3 \cdot 1\)

\(\displaystyle x \log 7 = 3\)

\(\displaystyle \frac{x \log 7}{\log 7} = \frac{3}{\log 7}\)

\(\displaystyle x = \frac{3}{\log 7}\)

Base-10 logarithms are very easy if the operands are a power of \(\displaystyle 10\).  Begin by rewriting the question:

\(\displaystyle \log{10000000}\)

Becomes...

\(\displaystyle \log{10^7}\)

because \(\displaystyle 10^7=10000000\)

Applying logarithm rules, you can factor out the \(\displaystyle 7\):

\(\displaystyle 7\log{10}\)

Now, \(\displaystyle \log{10}\) is \(\displaystyle 1\).

Therefore, your answer is \(\displaystyle 7\).

Base-10 logarithms are very easy if the operands are a power of \(\displaystyle 10\).  Begin by rewriting the question:

\(\displaystyle \log{100000}\)

Becomes...

\(\displaystyle \log{10^5}\)

because \(\displaystyle 10^5=100000\)

Applying logarithm rules, you can factor out the \(\displaystyle 5\):

\(\displaystyle 5\log{10}\)

Now, \(\displaystyle \log{10}\) is \(\displaystyle 1\).

Therefore, your answer is \(\displaystyle 5\).

Remember:

\(\displaystyle log_ab=c\) is the same as saying \(\displaystyle a^c = b\).

So when we ask "What is the value of \(\displaystyle log(1,000)\)?", all we're asking is "10 raised to which power equals 1,000?" Or, in an expression: 

\(\displaystyle 10^? = 1,000\).

From this, it should be easy to see that \(\displaystyle log(1000)=3\).

Without a subscript a logarithmic expression is base 10.

The expression \(\displaystyle \small \log(1000)=\log _{10}(1000)\) 

The logarithmic expression is asking 10 raised to what power equals 1000 or what is x when

\(\displaystyle \small 10^{x}=1000\)

We know that \(\displaystyle \small 10^{3}=1000\)

so \(\displaystyle \small \log (1000)=3\)

Rewrite the logarithm in division.

\(\displaystyle \log_{10}10^{3x+2} = \frac{\log{10}^{3x+2}}{\log 10}\)

As a log property, we can pull down the exponent of the power in front as the coefficient.

\(\displaystyle \frac{\log{(10)}^{3x+2}}{\log( 10)} = \frac{(3x+2)(\log 10)}{\log (10)}\)

Cancel out the \(\displaystyle \log(10)\).

The answer is:  \(\displaystyle 3x+2\)

When the base isn't explicitly defined, the log is base 10. For our problem, the first term

\(\displaystyle \log 10000=x\)

is asking:

\(\displaystyle 10^x=10000\)

\(\displaystyle x=4\)

For the second term,

\(\displaystyle \log(10)=y\)

is asking:

\(\displaystyle 10^y=10\)

\(\displaystyle y=1\)

So, our final answer is \(\displaystyle 4+1=5\)

By the reverse-FOIL method, we factor the polynomial as follows:

\(\displaystyle x^{2} -2x - 8 = (x - 4)(x + 2)\)

Therefore, we can use the property 

\(\displaystyle \ln ab = \ln a + \ln b\)

as follows:

\(\displaystyle \log \left ( x^{2} -2x - 8\right ) = \log [ (x - 4)(x + 2)] = \log (x-4) + \log (x + 2)\)

The first thing we can do is bring the exponent out of the log, to the front:

\(\displaystyle 3log(100)\)

Next, we evaluate \(\displaystyle log(100)\):

Recall that log without a specified base is base 10 thus 

\(\displaystyle \log_ba=c \Rightarrow b^c=a\)

\(\displaystyle \log_{10}100=c \Rightarrow 10^c=100 \Rightarrow c=2\).

Therefore

\(\displaystyle 3log(100)\)

becomes,

\(\displaystyle 3\cdot2\).

Finally, we do the simple multiplication:

\(\displaystyle 6\)