Solving equations is just a matter of doing stuff to both sides. You can usually do whatever you want to an equation as long as you do it to both sides and it isn't too weird (introducing division by zero, multiplying both sides by zero, etc). So when people start learning about exponential and logarithmic equations in high school, they're usually told that the log is the inverse of an exponent.
bx = a
can be freely converted to/from:
x = logba
So if you wanna solve an equation like 5x=7, all you have to do is take the log5 of both sides: log5(5x)=log57, which then becomes just x=log57. All of this comes naturally from the properties and identities of exponents and logarithms. You will do well to remember them (note, I use the division sign in place of fraction bars since it's easier to type, but just imagine that every ÷ is a fraction bar):
| Properties of exponents |
| axay = ax+y |
| ax÷ay = ax-y |
| (ax)y = axy |
| (ab)x = axbx |
| (a÷b)x = ax÷bx |
| a0 = 1 |
| a-x = 1÷ax |
| ax÷y = y√(ax) |
| Properties of logarithms |
| logb(xy) = logbx + logby |
| logb(x÷y) = logbx - logby |
| logb(ax) = xlogba |
| logb1 = 0 |
| logbb = 1 |
| blogba = a |
| logba = (logca)÷(logcb) |
| 1÷logba = logab |
And people will act like that's all there really is to them. And maybe it might as well be. They're sufficient enough to solve most high school exponential or log equations in a pretty easy way:
1. Solve 32x-3 = 6x
32x-3 = 6x
32x-3 = 3(log36)x
2x-3 = (log36)x
(2-log36)x = 3
3
x = -------
2-log36
not in ideal form yet, but whatever; comes out to ~8.129 where both sides will be ~2114602 or so
2. Solve ln(6x-5) = 70
ln(6x-5) = 70
eln(6x-5) = e70
6x-5 = e70
6x = e70 + 5
e70 + 5
x = -------
6
comes out to ~4.192x1029 (sorry) but it checks out
It's not too hard and it works. But look at the answer to #1. There's stuff you can do to it to make it look nicer. First of all,
imagine rewriting the 2 in the denominator as log3(32); that doesn't seem to do much, does it? Since the new exponent
of 2 can just come out as a coefficient, making it 2log33, and log33 is 1 according to one of the log rules. So
it's just 2(1) which is still 2.
But by turning it into a log, the denominator now looks like log3(32) - log36. And with another log rule, that can be rewritten as a single log: log3(9÷6), which simplifies to log3(1.5). Now the answer looks like:
3 --------- log3(1.5)And it still comes out to be the exact same number! But there's still more we can do. But that reciprocal log identity... what if we were to separate the fraction like:
3 1 --- * --------- 1 log3(1.5)That is a valid thing to do with a fraction. Then we can apply the reciprocal log identity, remembering that 3÷1 = 3:
3 * log1.53It still comes out to the same value! Now the coefficient of 3 can be put into the log........
log1.5(33)33 is 27, so the FINAL answer is log1.527. Amazing. It's like logs have a bunch of hidden identities that people don't want you to know about. And knowing them might be useful for solving the equation which this page is supposed to be about. So familiarize yourself with THESE, the log properties that the CIA doesn't want you to know:
| COOL Properties of logarithms |
| logbx = 1÷logxb |
| logb(1÷x) = -logbx |
| log(1÷b)x = -logbx |
| log(1÷b)(1÷x) = logbx |