Today's soundtrack is Spitfire: The Dead Next Door, a chaotic hardcore/screamo album from 1999, somewhat reminiscent of early Norma Jean. I really love the guitar work on the track "A Gaze at Quintessence;" it's slower, almost doomy, with some wicked pinch harmonics. The outro of "Good Cop, Bad Cop" is also crazy cool; it has some Herman Li-esque ascending riffs. I tried to find a YouTube vid of the songs to embed, but had no luck; however, the full album is in my blog playlist on Spotify.
This morning, I'm learning how to solve exponential and logarithmic equations.
We learned how to solve exponential equations in yesterday's lesson; let's review that now. Depending on whether the bases are exponentially related and on how many variables there are, there are two different methods that we can use.
If we have a single variable and a related base, our task is simple.
How to solve an exponential equation whose bases are exponentially related
Make the bases the same
If we start with 3²ˣ⁺¹ = 27ˣ, we can see right away that 27 is just 3³
3²ˣ⁺¹ = (3³)ˣ
Simplify exponents
3²ˣ⁺¹ = 3³ˣ
Drop the bases
2x+1 = 3x
Solve algebraically
2x-2x+1 = 3x-2x
1 = 1x
x=1
If the bases are not exponentially related, we must use the folllowing method:
How to solve an exponential equation with a single variable whose bases are unrelated
Apply the log to both sides
5ˣ = 20 becomes log5ˣ = log20
Apply the Power Rule to move all exponents down
xlog5 = log20
Divide both sides by the log and argument of the side with the variable
xlog5/log5 = log20/log5
Write out the answer
Exact form: x=(log20/log5)
Decimal form, rounded to five places: x=1.86135
But what if we have variables on both sides, and the bases are not related? Things get a lot messier, but it's still doable.
How to solve an exponential equation with variables on both sides whose bases are unrelated
Apply the log to both sides
4³ˣ⁺² = 3⁴ˣ becomes log4³ˣ⁺² = log3⁴ˣ
Use the Power Rule to move all exponents down
(3x+2)log4 = 4xlog3
Expand terms through distributive properties
3xlog4 + 2log4 = 4xlog3
Collect variables
3xlog4 - 4xlog3 = -2log4
x3log4 - x4log3 = -2log4
Factor to remove variable from terms
x(3log4 - 4log3) = -2log4
Isolate the variable through division
x = (-2log4/[3log4 - 4log3])
This is exact form
x = 11.76990
This is decimal form
We're now ready to move on to logarithmic equations. These equations start with a logarithm already present - no need to insert. Which method we use will depend on whether the variable and log are on the same side of the equation.
Let's go through some scenarios!
How to solve a logarithmic equation with log and variable on same side
Boot the log
If we start with logx = 2, we know that the invisible base is 10
"Boot the log" by sliding the base to the other side of the equation and making the constant into its exponent, which gives us x = 10²
Simplify
10² = 100; thus, x = 100.
How to solve a logarithmic equation with log and variable on opposite sides
Use the Change of Base rule
log₄21 = x + 1 becomes log21/log4 = x+1
Isolate the variable
(log21/log4) -1 = x
Write in either exact form or decimal form
To sum this all up: when solving an equation involving a single log, if the log and variable are together, boot the log. If the log and variable are apart, change the base.
We can use the laws of logarithms to solve more complex logarithmic equations.
How to solve a logarithmic equation with three logs
logx - log4 = log5
logx - log4 + log4 = log5
logx = log4 + log5
logx = log(4+5)
Now, remembering that two logs added together indicate multiplication of the arguments...
logx = log20
Since we have log[argument] on both sides, we can drop the logs
x = 20
How to solve a logarithm with two logs
2logx = log81
Remembering the power rule, we know that the 2 preceding "log" on the left indicates that x is raised to a power of two
logx² = log81
81 is a perfect square; let's change its base
logx² = log9²
We now have log[argument] on both sides; let's drop the logs
x² = 9²
Take the square root of both sides
x = ±9
Remember the restrictions on our values: x>0, and x≠1, so -9 is an extraneous root.
x = 9
How to solve a logarithm that turns into a polynomial
log₄(-x) + log₄(2-x) = log₄15
Using the multiplication law, we know that adding two logs with the same base together multiplies them, so on the left side, after using distributive factors, we get log₄(-2x+x²)
log₄(-2x+x²) = log₄15
Now we can drop the logs
-2x+x²=15
Move all onto one side and solve as a quadratic polynomial for x
x²-2x-15=0
This factors into (x-5)(x+3)
x = -3 and x=5; however, notice that in the original equation, our variables were preceded by negative signs, so if we use a positive argument, we will get a negative log, which is illegal. So our answer is -3.
That's all for today; tomorrow, we'll be talking about exponential functions.