Subtracting Big Powers

$\large \dfrac{6^{2007} - 6^{2006}}{30} = \dfrac{6^{2006} (6 - 1)}{30} = \dfrac{6^{2006} \cdot 5}{30} = \dfrac{6^{2006}}{6} = 6^{2005}$

1st step we can change $6^{2007}$ to become : $6^{2006} \times 6$ so, $\frac {6^{2006}(6) - 6^{2006}}{30}$ $\frac {6^{2006} (6-1)}{30}$ $\frac {6^{2006}(5)}{30}$ $\frac {6^{2005}(5)(6)}{30}$ $\frac {6^{2005}(30)}{30} = 6^{2005}$

My way :

$6^{2007} = 6 \times 6^{2006}$

So:

$6^{2007} -6^{2006} = 6 \times 6^{2006} - 6^{2006} = 5 \times 6 ^{2006}$

And:

$6 ^{2006} = 6 \times 6^{2005}$

Now is easy:

$6^{2007} -6^{2006} = 5 \times 6 \times 6^{2005} = 30 \times 6^{2005}$

$\frac{6^{2007} -6^{2006}}{30} = \frac{30}{30} \times 6^{2005} = 6^{2005}$

The trick is solving the behavior in the numerator first. Consider the problem with smaller numbers:

$6^{3} - 6^{2} = 216 - 36 = 180$ Here we see that if we did 6 cubed minus 6 squared we would get 6 cubed - 36 (makes sense because 6^3 = 6 * 6 = 36 * 6) Recall 6 * 36 = 36 + 36 + 36 +36 + 36 + 36 So what we have done essentially is subtracted one term from the final number This gives us the following equation $6^{3} - 6^{2} = 6^{2}*5 or (6 - 1)$

Now that we solved a trivial example, we can apply it to all cases. $\frac{6^{2007}-6^{2006}}{30} = \frac{6^{2006} * (6 - 1)}{30} = \frac{6^{2006}*5}{30}$ $= \frac{5}{30} *6^{2006} = \frac{6^{2006}}{6} - //reduce-the-fraction$ $\frac{6^{2006}}{6} = \frac{6^{2006}}{6^{1}} = 6^{2006-1} = 6^{2005}//Laws-of-exponential-math$

This process of problem solving is known as induction and can be used to solve most solutions by finding a pattern in smaller trivial cases

6^(2007)- 6^(2006)/30
= 6^(2006) (6-1)/30
= 6^(2006)
5/30
= 6^(2006)/6
= 6^(2005)

$\frac{6^{2007}-6^{2006}}{30}=\frac{6^{2005} \times 6(6-1)}{30}\\ =\frac{6^{2005}\times 30}{30}\\ =6^{2005}$

I was doing a freaking mess with logaritms for at least one hour when I realized everything.

$\frac{6^{2007}-6^{2006}}{30}=...$ $...=\frac{6(6^{2006})-6^{2006}}{30}=...$ $...=\frac{5(6^{2006})}{30}=\frac{6^{2006}}{6}=...$ $...=6^{2005}$

$\frac{6^{2007}-6^{2006}}{30} = \frac{5 \cdot 6^{2006}}{5\cdot 6} = 6^{2005}$

(6^2007-6^2006)/30=(6*6^2006-6^2006)/30={6^2006(6-1)}/30=(6^2006 *5)/30=6^2006/6=6^2005

What I did was to first simplify the fraction:

$\frac{6^{2007}-6^{2006}}{6\cdot5}$ $=$ $\frac{6^{2006}-6^{2005}}{5}$

Then I started thinking about ways to solve this problem and instead of factorizing I observed a pattern in the differences of the powers of $6$ :

$6^2-6=\boxed{30}$

$6^3-6^2=\boxed{180}$

$6^4-6^3=\boxed{1080}$

$6^5-6^4=\boxed{6480}$

$...$

As you can see the difference of two consecutive powers of $6$ increases by a factor of six as the degrees increase by one:

$30, 180, 1080, 6480, ...$

And I realized that the first equation was just the multiplication $6\cdot5=30$ and if I applied this concept to the other equations it would be as follows:

$6^3-6^2=6^2\cdot5$

$6^4-6^3=6^3\cdot5$

$6^5-6^4=6^4\cdot5$

$6^{x+1}-6^x=6^x\cdot5$

So now we substitute $x$ :

$\frac{6^{2006}-6^{2005}}{5}$ $=$ $\frac{6^{2005}\cdot5}{5}$

$\frac{6^{2005}\cdot5}{5}$ $=\boxed{6^{2005}}$

$\begin{aligned} \frac { { 6 }^{ 2007 }-{ 6 }^{ 2006 } }{ 30 } & = & \frac { { 6 }^{ 2006 }\cdot 6-{ 6 }^{ 2006 } }{ 30 } \\ \quad & = & \frac { { 6 }^{ 2006 }(6-1) }{ 30 } \\ \quad & = & \frac { 5({ 6 }^{ 2006 }) }{ 5\cdot 6 } \\ \quad & = & \frac { { 6 }^{ 2006 } }{ 6 } \\ \quad & = & \boxed { { 6 }^{ 2005 } } \end{aligned}$

$\frac{6^{2007}-6^{2006}}{30}=\frac{6^{2006}(6 - 1)}{30}=\frac{6^{2006} \times 5}{30}=\frac{6^{2006}}{6}=6^{2006-1}=\boxed{\large{6^{2005}}}$

$\frac{6^{2007} - 6^{2006}}{30}$ = $\frac{6^{2006}(6 - 1)}{30}$ = $\frac{6^{2006} \cdot 5}{30}$ = $\frac{6^2006}{6}$ = $6^{2005}$

Common factor top

Reduce the quotient to show

Six to the power 2005

We can take 6^(2006) as a common

value..i.e,

6^(2006)(6-1)/30

Now, we can solve it as..

6^(2006)×5/30

After solving further,we get

6^(2006)/6

which equals to 6^(2005)

which is the answer.