Underlying Reason For Taking Log Base 10 For the equation $2^x = 7$
The textbook says to use log base ten to solve it like this $\log 2^x = \log 7$. 
I then re-arrange it so that it reads $x \log 2 = \log 7$ then divide the RHS by $\log 2$ to isolate the $x$. I understand this part.
I can alternatively solve it in an easier way by simply using $\log_2 7$ on my calculator.
Using both methods the answer comes to the same which is $2.807$
My question is twofold:


*

*Why would the textbook suggest to use log base ten rather than simply using log base two?

*I can see how using log base ten and the suggested method in the textbook makes me arrive at the same answer but I don't understand WHY this is so. How does base ten play a factor in the whole scope of things.
Thank you
 A: This works in any base because $\log_b (a^c) = c \log_b(a)$ 
The practical reason for using base $10$ was a little old fashioned: it allowed the use of tables of logarithms instead of a calculator and reducing these calculations to addition and subtraction.  Calculators then provided a $\log_{10}$ function as a carry-over from tables and just called the button log
My 1953 elementary statistical tables have logarithms of numbers from $1$ to $10$ and anti-logarithms of numbers from $0$ to $1$.  Since these are logarithms base $10$, I can easily deal with all numbers because $\log(a \times 10^n)=n +\log(a)$
Here I want $\dfrac{\log 7}{\log 2}$.  


*

*My tables tell me $\log 7\approx 0.8451_6$ (with the ${\,}_6$ helping interpolation) and $\log 2\approx 0.3010_{22}$.  So that leaves me with trying to calculate $\dfrac{0.8451}{0.3010}$. I cannot be bothered to do long division, so instead I try to calculate $\text{antilog}\,({\log 0.8451 -\log 0.3010})$. 

*My tables tell me $\log 8.45 \approx 0.9269_5$ so I write $\log 0.8451 \approx \bar{1}.9269$ (with the $\bar{1}$ because I wanted $\log (8.451\times 10^{-1})= -1+\log 8.451$).  Similarly it tells me $\log 3.01 \approx 0.4786_{14}$ so I write $\log 0.3010 \approx \bar{1}.4786$

*I now calculate by hand $\bar{1}.9269 - \bar{1}.4786 = 0.4483$.  My tables tell me $\text{antilog}\,0.448 \approx 2.805_7$ and I interpolate the final $3$ using the ${\,}_7$ to give a final approximate answer of $2.807$.  Which is what you got with some clever silicon
A: The use of Log in base 10 is a common choice and we often find it in calculators since our standard system of counting is in base 10 (i.e. number of fingers).
To handle algebraic problems any other choice is fine depending upon the specific problem to solve by logarithmic identities.
For the application in calculus the best choice is the natural logarithm in base e.
A: Other answers mention the change of base $$ \log_a x = \frac {\log_b x}{\log_b a}. $$ An immediate consequence is that a quotient of logarithms does not depend on the base: $$\frac {\log_ax}{\log_ay}=\frac {\frac {\log_bx}{\log_ba}}{\frac {\log_by}{\log_ba}}
=\frac {\log_bx}{\log_by}. $$
A: As others have said, the formula works for a logarithm of any base, because of the change-of-base formula. However, the accepted answer says, “The practical reason for using base 10 was a little old fashioned: it allowed the use of tables of logarithms instead of a calculator and reducing these calculations to addition and subtraction.”  This raises the question, why did those tables use base 10?  After all, we just explained that the same tricks work for any base at all.  If you just needed to pick one base to compile a table of and print as a book, or mark on a slide rule, wouldn’t the least-arbitrary choice have been e?
The answer behind the answer is that base-10 logarithms are the easiest for humans without calculators and used to decimal numbers to work with.  The log of 1 is 0, the log of 10 is 1, the log of 100 is 2, and so on up, so any one-digit number has a log of zero point something, any two-digit number has a log of one point something, and so forth.  So, without a calculator, what’s the log of 300?  Well, log 300 is log 3·100, which is log 3 + log 100.  The square root of 10 is a little more than 3, so log 3 is a little less than 0.5, and log 100 is exactly 2, so intuitively it’s a little less than 2.5.  That makes it really easy to find the log of any number in scientific notation, or move a decimal point left or right.
If you did a lot of these, which engineers once had to, you’d quickly get an intuitive sense for it.  And doing these problems with base 10 lets you use that number sense to catch silly mistakes: “No, that can’t be right, it’s something thousand, so the log has to be three point something.”
A: The only reason we use base ten for anything (including decimal numbers) is that biological evolution endowed humans with ten fingers. The Mayans are supposed to have used base twenty, apparently because they were barefoot and thought that toes were convenient for counting too.
A: Another common type of question in logarithms is
$$2^{x+2} = 3^{2x}$$
Now suppose you have learnt always to use $\log_2$ if the base is 2, and $\log_3$ if the base is 3. You are now stuck! Which base should you use?
On the other hand suppose you have learned always to use $\log_{10}$, then you get
$$(x+2)\log2 = 2x\log3$$
which you can solve. It is important that you know the $\log(a^b) = b\log(a)$ rule, and solving using base 10 helps you to practice this method.
So a student who learns only one technique for solving equations with logarithms can learn to use base 10. Better students can then learn to use different bases, if another base is more convenient. This is why the textbook may teach you to use base 10: It is more general and practices an important formula.
As for why "10", as other answers have noted, any base is possible, but base 10 is convenient since decimal numbers are based around 10.
A: Question 1) As you see, both method arrive at the same answer, however $\log$ base 10 is a natural choice, as stated in the comment by  Matti P above.
Question 2) It is because for $a>0,a\neq 1,b>0,  c>0, c\neq 1$,$$\log_a b=\frac{\log_c b}{\log_c a}$$In particular in your example, $c=10$, so both method arrives the same answer. The formula is known as the change of base formula, see here.
A: Logarithmic functions enjoy many properties.
One of the very interesting properties of logarithms is the formula called change of base formula.
$$ \log_a x = \frac {\log_b x}{\log_b a} $$
For example $$ \log_2 7 = \frac {\log_{10} 7}{\log_{10} 2} $$
This formula makes finding logarithms in an arbitrary base possible by using only logarithms base $10$ or natural logarithms.   
A: 
Why would the textbook suggest to use log base ten rather than simply using log base two?

Hmmm, if we embrace binary, and use log2(), how about using base 2 for the constants and representation also?  
Then $2^x = 7$ becomes $10^x = 111$ and is solved with $x = 10.1100111010...$
When numbers are represented in base 10. Using logn() other than $n==10$ takes an unnecessary additional manipulation and prevents observing niceties like log2(100) == 2.


How does base ten play a factor in the whole scope of things. (?)

As base 10 is common understood by the target audience (students), using log10() is also more readily understood by them. It is more instructive than employing base 2.
On other hand, if the textbook was for processors, using base 2 makes more sense.
