How do you solve Exponential Equations?

[Abstract_Logic]

I'm learning College Algebra. I'm wondering how to solve Exponential Equations, such as

 

8^(X^2 - 2X) = 1/2

 

8 to the [X squared minus 2X]th power = 1/2

 

and other equations of that nature.

[Cap'n Refsmmat]

Logarithms!

 

[math]8^{x^2 - 2x} = 0.5[/math]

 

[math]\log 8^{x^2 - 2x} = \log 0.5[/math]

 

[math](x^2 - 2x) \log 8 = \log 0.5[/math]

 

[math]x^2 - 2x = \frac{\log 0.5}{\log 8}[/math]

 

and you can see how that will work out in the end.

 

http://en.wikipedia.org/wiki/Logarithm

[Bignose]

The other trick is to raise everything to the exact same power and then equate powers.

 

i.e. [math]\frac{1}{2} = 8^c[/math]. Figure out what power 8 needs to be raised to to equal one half and then equate what is in the exponent on the LHS to what is in the exponent in the RHS. Both methods give the same answer, and actually both methods use the same idea (you will use a logarithm to find what power to raise 8 to to equal 1/2 whether you knowingly do it or not).

 

That said, I think that the logarithms route is the better one to know, because logarithms have several useful relations that typically make these problems easier.

[Shadow]

Logarithms!

 

[math]8^{x^2 - 2x} = 0.5[/math]

 

[math]\log 8^{x^2 - 2x} = \log 0.5[/math]

 

[math](x^2 - 2x) \log 8 = \log 0.5[/math]

 

[math]x^2 - 2x = \frac{\log 0.5}{\log 8}[/math]

 

and you can see how that will work out in the end.

 

http://en.wikipedia.org/wiki/Logarithm

 

One thing worth knowing is that you can choose the base of the logarithm, ie. instead of getting [math]\log 8^{x^2 - 2x} = \log 0.5[/math] you can get [math]\log_2 8^{x^2 - 2x} = \log_2 0.5[/math]. So then in Capn's last step, instead of [math]x^2 - 2x = \frac{\log 0.5}{\log 8}[/math] you get [math]x^2 - 2x = -\frac{1}{3}[/math]. Alternatively you can also get this result by using [math]\frac{\log a}{\log b} = \log_b a[/math]. So [math]\frac{\log 0.5}{\log 8} = \log_{8} \frac{1}{2}=-\frac{1}{3}[/math].

 

Just my two cents

[Petanquell]

I thing Big Nose said it perfect.

 

[math] 8^{x^2 - 2x} = \frac{1}{2} [/math]

the first step is:

[math] 8^{x^2 - 2x} = 8^{-\frac{1}{3}}[/math] using fact that [math]\frac{1}{2}=8^{-\frac{1}{3}}[/math]

Than using the old rule "when bases are equal than exponents must be equal too" or with "more official" way, use logarithm with base 8 on both sides we get:

[math]\mathrm{log_{8}}(8^{x^2 - 2x})=\mathrm{log_{8}}(8^{-\frac{1}{3}}) [/math]

[math](x^2 - 2x) \cdot \mathrm{log_{8}}(8)=-\frac{1}{3} \cdot \mathrm{log_{8}(8)} [/math]

and we get to easy equation

[math] x^2 - 2x= -\frac{1}{3} [/math]

 

or in the first step use that [math](\frac{1}{2})^{-3}=8[/math] and do the same again only with some small changes:

 

[math] ((\frac{1}{2})^{-3})^{x^2 - 2x} = \frac{1}{2}[/math]

[math]\mathrm{log_{\frac{1}{2}}}((\frac{1}{2})^{-3})^{x^2 - 2x}=\mathrm{log_{\frac{1}{2}}}\frac{1}{2} [/math]

[math]\mathrm{log_{\frac{1}{2}}}(\frac{1}{2})^{-3x^2 + 6x}=\mathrm{log_{\frac{1}{2}}}\frac{1}{2} [/math]

[math](-3x^2 + 6x) \cdot \mathrm{log_{\frac{1}{2}}}\frac{1}{2}=\mathrm{log_{\frac{1}{2}}}\frac{1}{2} [/math]

 

[math] -3x^2 + 6x= 1 [/math] (dividing with [math]-3[/math] you get the same equation as before..)

 

 

I hope it helps.

 

Pq