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math:log_exp [2020/12/14 02:14] – Added inverse property epixmath:log_exp [2022/01/24 15:36] (current) – fixed missing base epix
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 This property applies to equations that look like $a^x = a^y$. If $x = y$, then you can drop the exponent. The same works for logs (i.e. $\ln x = \ln y$ -> $x = y$). This property applies to equations that look like $a^x = a^y$. If $x = y$, then you can drop the exponent. The same works for logs (i.e. $\ln x = \ln y$ -> $x = y$).
 ==== Inverse Property ==== ==== Inverse Property ====
 +This one is pretty self explanatory.
   * $a^{\log_{a}x} = x$   * $a^{\log_{a}x} = x$
   * $\log_{a}a^x = x$   * $\log_{a}a^x = x$
 +==== Multiplication Property ====
 +The log of a product is the sum of the logs.
 +  * $\log_{a}xy = \log_{a}x + \log_{a}y$
 +
 +==== Division Property ====
 +The log of a quotient is the difference of the logs.
 +  * $\log_{a}\frac{x}{y} = \log_{a}x - \log_{a}y$
 +
 +==== Log to a Power Property ====
 +The exponent on the argument is the coefficient of the log.
 +  * $\log_{a}x^r = r\log_{a}x$
 +
 +===== Change of Base Formula =====
 +While the TI-84 Plus CE has a $\log_{base}$ function((math -> A: logBASE()) that exempts you from needing this for most problems, it is still good to know and even required if you use an older, shittier, calculator like the normal TI-84 Plus.
 + The formula goes as follows: $\log_{c}a = \frac{\log_{b}a}{\log_{b}c}$ where $a$ is the result, $b$ is the new base, and $c$ is the old base. As a practical example, with the TI-84, the only $\log$ function available is the common log, or $\log_{10}$(($\log$ without a base specified is base 10)), so to calculate values with the TI-84 of different bases, you have to use the change of base formula to base 10.
 +
 +===== Quadratic Exponentials =====
 +While these types of problems look daunting((and typically contain $e$ to scare poor highschool precalc students)), they are literally the same as solving a quadratic. Take $e^{2x}-4e^x-5=0$ for example. All you need to do is take out the square with the [[math:exponential_rules|Exponential Rules]]. So the equation above would become $(e^x)^2-4e^x-5=0$, and there you see the quadratic more clearly((still don't? swap $e^x$ with a variable like $x$ -> $x^2-4x-5=0$ where $x=e^x$)). Then, solve the quadratic as usual.

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