**Hey there š** I would like to quickly plug a product I am working on to eliminate blind spots before they become problems. If you are building a software product and want to know when your leads are about to churn, serve customers better and close more deals, make sure to check it out!

Preparing for my exams around Statistics and Macroeconomics, I've been using a couple of rules a lot for transforming and solving equations. When they were first introduced, I didn't attempt to memorize them but after having spent a great time going through exercises, I've changed my mind.

Note that this post will focus heavily on exponent, radical, and fraction rules, less so on other topics. I'll also give a typical setting for each rule and how it would be applied in such a case.

This can be helpful if a already is a fraction. In this case, you can easily multiply both the numerator of a with 1 (which will stay the same) and the denominator of a with b.

Example: ${{1 \over 2} \over {3 \over 4}}= {1 \over 2}* {1 \over {3 \over 4}}= {1 \over 2} * {4 \over 3}= {4 \over 6}$

One of the simpler rules to simplify terms is that if the same base is used, you can rewrite it to using the base once and sum up all exponents.

Example: $a^3+a^2=a^5$

One of the simpler rules is that if you have a fraction with the same exponents for both numerator and denominator, you can simplify it to adding the exponent just once, for the complete fraction.

Example: ${a^2 \over b^2}=({a \over b})^2$

What you'll often come across is the same base in a fraction, but with different exponents. You can combine it by subtracting the exponent of the denominator from the exponent of the numerator.

Example: ${a^5 \over a^3} ={a^{5-3}}=a^{2}$

If you need a different way to transform the fraction from above and the exponent in the denominator is larger than the exponent in the numerator, you can rewrite it this way.

You can simplify stacked fractions by multiplying the numerator and denominator of the lower part and dividing by the denominator. If your numerator is 1, you simply remove it and flip the lower fraction to get your result.

If you want a more extreme example, ${{a \over b} \over {c \over d}} = {{a *d} \over {b*c}}$ works too.

Example: ${5 \over {7 \over 9}}={5*9 \over 7}$

Negative exponents might sound scary but refer to your base being the denominator of a fraction dividing one. If your exponent is not -1, don't forget to add it to the denominator in your fraction.

Example: $5^{-3}={1 \over 5^3}$

Using our rule to simplify fractions with the same exponent, then applying the rule of negative exponents, followed by stacked fractions, we get to a swapped representation with a positive fraction in the end. Calculating this is much simpler.

Example: ${5^{-2} \over 7^{-2}}={({5 \over 7})^{-2}}={1 \over {5 \over 7}^{2}}=({7 \over 5})^2$

Square roots can be represented simplify by a fraction of $1 \over 2$. In general, all roots can be represented as fractions and vice-versa, making it easy to use with other rules on this page, as you can perform the same transformation steps.

${\sqrt{3} \over \sqrt{2}}={{3^{1 \over 2}} \over 2^{1 \over 2}} = {3 \over 2}^{1 \over 2} = \sqrt{1.5}= 1.224 \dots$

Combining the rule about roots with negative exponents, we can simplify any fraction over a root. This also works in a generic way, if the numerator is not 1: ${1 \over \sqrt[c]{a^b}} = a^{-{b \over c}}$.

Example: ${1 \over \sqrt{4}}=4^{-{1 \over 2}}$

If we have multiple layers of exponents, we can combine them by multiplying the exponents.

${2^2}^2=2^{2*2}=2^4=16$

If you ever come across the question of how to get rid of an exponent, we need to look into the multiplicative inverse or reciprocal. "Removing" an exponent can only be done by transforming it to 1, e.g. $a^x=...=a^1=a$. We can do this by using the rule from above, nesting the term in another exponent which is the inverse: $(a^x)^{1 \over x}=a^{x*{1 \over x}}=a^1=a$.

Example: Transform to $a, a^{-2} =({a^{-2}})^{- {1 \over 2}}=a^{-2 * -{1 \over 2}}=a^1=a$

A similar idea works for the term $\sqrt[n]{a^n}=a$ as $\sqrt[n]{a^n}=a^{n \over n}=a^{1}=a$

This might be helpful in some cases where you're working with fractions already. This works because $a^b= {1 \over a^{-b}}={1 \over {1 \over a^b}}={{1*a^b} \over 1}=a^b$.

Example: $5^3={1 \over 5^{-3}}$

I'll update this whenever I use an additional rule for working with fractions and exponents. If you have any questions, suggestions, or feedback in general, don't hesitate to reach out via mail or Twitter.

**Thanks for reading this post š** I would like to quickly plug a product I am working on to eliminate blind spots before they become problems. If you are building a software product and want to know when your leads are about to churn, serve customers better and close more deals, make sure to check it out!

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