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

Question

Answer 1

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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.

 

Answer 2

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?

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