Monday, November 9th: change of base and solving exponential equations.

Today we learned about the change of base formula; also, it was semester day 40, so we are halfway in. Mr. Max believes that we should work hard at our accelerated math and do the relevant sheets before the test. If we were to do that, we'd do much better on the test. If you're a little behind on the sheets for our upcoming test, that's fine. You could always get caught up before the next test, and keeping on top of accelerated math would make things less stressful.

logbn = logan / logab

As long as we follow the pattern of the formula, you can make it workable on any base you choose.

given y = log2 3

can be rewritten as:

y = log10 3 / log10 2

How to we evaluate? We can either use our calculator and get y~ 1.585 . . .

Or, we can use logs. (No kidding, eh!)

y = log2 3

change to an exponential expression.

2^y = 3

then we apply the def of logs: log(2^y) = log3

y = log10(3) / log10(2)



Solving exponential equation.

Solve:

2(3^x) = 5
CAVEAT! (A latin word that means "warning" or "beware"):
log (x + 3) != (not equal to) logx + log3
instead, it becomes:
log(2(3^x)) = log5
log2 + log3^x = log5
log2 + x * log3 = log5
x * log3 = log5 - log2
x = log5 - log2 / log3 (at this point, we cannot continue without a calculator)
x = .8340437671
to check out answer: 2 * 3 ^ .8340437671 = 5

We just solved our very first exponential equation!




Let's solve another one:

19^(x - 5) = 3^(x+2)
(Note: Our reason for doing the log is to bring the exponents down.)
log 19 ^ (x - 5) = log 3^(x+2)
x * log19 - 5 * log19 = x * log3 + 2 * log 3
(x - 5) * log 19 = (x + 2) * log 3
x * log19 - 5 * log19 = x * log3 + 2 * log3
x * (log19 - log3) = 2 * log3 + 5 * log19
x = 2 * log3 + 5 * log19 / (log19 - log3) (now we use our calculator)
x ~ 9.166




log2 (x - 2) + log2(x) = log23
log2(x - 2)(x) = log23
x - 2(x) = 3 (we can remove the logs since they have the same base)
x ^ 2 - 2x - 3
(x + 1)(x - 3) = 0
x = -1, x = 3

STOP!

x = -1 doesn't solve the equation.

x = 3 works, though




log5(3x + 1) + log5(x - 3) = 3
log5[(3x+1)(x-3)] = 3

(use definition of logs:
5 ^ 3 = (3x + 1)(x - 3)
logab = n
a^n = b)

125 = 3x^2 - 8x -3
0 = 3x^2 - 8x - 128
(384: 16 and 24 as factors)
0 = 3x^2 - 24x + 16x - 128
0 = 3x(x-8) + 16(x-8)
0 = (3x + 16) (x - 8)
3x + 16 = 0, x - 8 = 0
x = -16/3 and x = 8

check for error:
- 16 /3 isn't a solution because after we plug it in, there is no value that 5 ^x that would be equal to a negative argument.