Solving Exponential and Log Equations:

Hi guys! During the last couple of class periods, we have learned how to solve equations using logs. We learned two specific methods.

The First Method:

Example:        2^x = 35

Step one - we just rewrote the equation as a log:

              

log2(<the base)35 = x

Step two - divide the log of 35 (what your original equation is equal to) by the log of 2 (the base):

log 35  

 log 2   = X

the answer for this example when done out is 5.12928, but Lisa has asked us to leave the exact answer which is:    log 35  

                   log 2 

*to help you remember that the base becomes the denominator, just look at the equation rewritten as a log, and the base is lower than the number that stands for what your original equation was equal to.

The Second Method:

Example:        2^x = 35  

Step one - take the log of both sides of the equation:

log 2^x  = log 35 

Step two - use the power rule, which allows you to bring any power in the log (x) out in front of it:

xlog 2 = log 35

Then solve by...

x =  log 35  

        log 2

*remember logs are not variables

We also learned how to simplify log equations using the Properties of Logarithms (which can be seen on Jessie's blog [the one before mine..]) 

Here's an example from our homework that we went over in class:

loga (m^8n^12) ^1/4

           a^3b^5

first we use the power rule:

1/4 loga (m^8n^12)

                 a^3b^5

then we use the quotient rule:

1/4 [loga (m^8n^12) - loga (a^3b^5)]

then we used the product rule:

1/4 [loga m^8 + loga n^12 - (loga a^3 + loga b^5)]

then we used a property in the category of other, as well as the power rule:

1/4 [8 loga m + 12 loga n - (3 + 5 loga b)]

we then just multiplied the 1/4 through the rest of the equation:

2 loga m + 3 loga n - (3/4 + 5/4 loga b)]

Hope this helps!

-Timmy