# [Pre-calc] Answer these two questions and show work pls?

### 2 Answers

- stanschimLv 72 months agoFavourite answer
1. (3)^(x/5) + (3)^[(x-10)/10] = 84

(3)^(x/5) + (3)^[(x-10)/10] = 3^4 + 3^1

x = 20 satisfies both conditions.

2. (3^2)(3^5)(3^8)...(3^(3n-1)) = 27^5

(3^2)(3^5)(3^8)...(3^(3n-1)) = (3^3)^5

(3^2)(3^5)(3^8)...(3^(3n-1)) = 3^15

Using just the first three terms satisfies this equation.

3^(2+5+8) = 3^15

Therefore, n = 3.

- ?Lv 72 months ago
1) hint: let u= 10x thus you get (3²)^u + 3^u / ( 3^(1/10))^10 = 84. Thus,

9^u + 3^u / 3 = 84. Thus can quickly be solved by inspection (for the trivial solution...)

2) Hint: 27 = 3^3, thus 27^5 = (3^3)^5 = 3^15.

Now back to the LHS: The sum of the exponents must give 15. So, when do you stop the sequence?

Dont forget to vote me best answer for being the first to correctly walk you through and especially for not spoiling out the answer!