MHA4: Solving Exponential Equations

Logarithms are quite handy, because not only they are useful when it comes to solving logarithmic equations, but also when it comes to solving exponential equations. Exponential equations aren't something to be worried about. Exponential equations are equations where the variables are in the exponents. How come logarithms are used to solve exponential equations? Let's have an example.

Example 1:
Solve for x: 6^(x + 4) = 3^(x - 3)

We can't do anything much with the x's, let alone solve for its value, unless we can end their days as an exponent. This is why logarithms are used here. Remember property #1? This property can solve our problem! Taking the logarithm of both sides:

log 6^(x + 4) = log 3^(x - 3)
(x + 4)log6 = (x - 3)log3 (Property #1)
xlog6 + 4log6 = xlog3 - 3log3
xlog6 - xlog3 = -4log6 - 3log3
(log6 - log3)x = log(6⁻⁴) + log(3⁻³) (Property #1)
xlog2 = log(1/1296) + log(1/27) (Property #3)
xlog2 = log(1/34992) (Property #2)
x = log(1/34992) / log2
x = log₂(1/34992) (Property #5)
x ≈ -15.0099

Example 2:
Solve for x: 7^(x – 2) = 5^x

I won't put the property "thingy" anymore. I'm pretty sure you're familiar with the properties now. Let's proceed to the solution.

Solution:
7^(x – 2) = 5^x
log 7^(x – 2) = log 5^x
(x - 2)log7 = xlog5
xlog7 - 2log7 = xlog5
xlog7 - xlog5 = 2log7
(log7 - log5)x =log(7
xlog(7/5) = log(7²)
x = log49 / log(7/5)
x = log_7/549
x ≈ 11.5665

Test your skills by solving these problems!
1. Solve for x: (1/4)^(x + 1) = 16ˣ Answer: -1/3
2. Solve for x: 2.3^(2x + 1) = 9.1/(0.83^(8x))
Answer: x = (log(9.1/2.3))/(log(2.3^2 * 0.83^8)) or 7.8511