Solving exponential equations like $6^{3x}=4^{2x-3}$ using logarithms

Question

I'm trying to solve these using logarithms:

$a$) $9^{x+1} = 27^{2x-3}$

$b$) $6^{3x} =4^{2x-3}$

$c$) $210=40(1.5)^x.$

---

I'm trying to practice logarithms by doing various questions. It's been a while since I've done them, so I was hoping to refresh on them a bit. Hoping someone can walk me through the steps.

I'm not sure how to go about a, so since a and b are so similar I will probably have difficulty with both. C I Wasn't sure if I should start with finding a common base (since I could get a common base of 2 with them without too much difficulty) first or if I should go ahead and take the log of both sides. I'm trying to think reverse logarithm laws as well, but for these questions I'm just not sure of what I should do first. If I see how to do a or b I should be able to figure out the other. But C I may need some guidance

Answer 1

Hints: $$\log(a^b) = b \log(a) \\ \log(ab) = \log(a) + \log(b) $$

Example for part $c)$:

$$ 210 = 40(1.5)^x \\ \log(210) = \log(40 (1.5)^x) \\ \log(210) = x\log(1.5) + \log(40) \\ \frac{\log(210) - \log(40)}{\log(1.5)} = x \\ x \approx 4.0897 $$

Answer 2

In reference to Axoren, the following is true for (1): \begin{align} 9^{x+1}=27^{2x-3} & \implies\left(3^2\right)^{x+1}=\left(3^3\right)^{2x-3}\tag{1}\\[1em] & \implies \left(3\right)^{2x+2}=\left(3\right)^{6x-9}\tag{2} \\[1em] & \implies \left(2x+2\right)\log_3\left|3\right|=\left(6x-9\right)\log_3\left|3\right|\tag{3}\\[1em] & \implies 2x+2=6x-9\tag{4} \\[1em] & \implies \boxed{x=\frac{11}{4}.}\tag{5} \end{align} I will leave the rest to you.