Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to find the value of x in the following problem, I have to solve it without logarithm.

Problem : $$ \dfrac {27 ^ {(2x+1)} } { 3 ^ {(x+1){5}}} = \dfrac{1}{3} $$

EDIT: My work so far:

$$ \dfrac {3^{3(2x+1)} } { 3 ^ {(x+1)5}} = 3^{-1} $$

I know the formula $ b^{u} = b^{v} <=> u = v $ but I am not able to use it with this problem.

Thanks for help !

share|improve this question
    
Hint: $27=3^3$. –  David Mitra Nov 11 '11 at 18:43
    
$$ \frac{3^{3(2x+1)}}{3^{5(x+1)}}=3^{-1}$$ –  pedja Nov 11 '11 at 18:45
    
Additionally, $\frac{3^a}{3^b}=3^{a-b}$ and $\frac1{3^a}=3^{-a}$... –  J. M. Nov 11 '11 at 18:45
add comment

1 Answer 1

up vote 3 down vote accepted

Let's combine all the hints (from above comments) to write:

$\dfrac{27^{(2x + 1)}}{3^{5(x + 1)}} = \dfrac{(3^3)^{(2x + 1)}}{3^{(5x + 5)}} = \dfrac{3^{3(2x + 1)}}{3^{(5x + 5)}} = \dfrac{3^{(6x + 3)}}{3^{(5x + 5)}} = 3^{(6x + 3) -(5x + 5)} = 3^{-1}$

share|improve this answer
    
David gives us the first equality, pedja gives us the second, and J.M. gives us the final two. Can you take it from here? –  The Chaz 2.0 Nov 11 '11 at 18:51
1  
Is it $$ 6x+3-5x-5 = -1, x = 1 $$ ? –  Jonathan Nov 11 '11 at 18:57
    
Come on, now! You have to distribute that negative/minus!! So it would be $6x + 3 - 5x -5 = -1$ ... –  The Chaz 2.0 Nov 11 '11 at 18:58
    
sorry! I've just edited it :) –  Jonathan Nov 11 '11 at 18:59
    
There you go! And you can always check your solution by plugging $x = 1$ into the original equation. And welcome to math.stackexchange! –  The Chaz 2.0 Nov 11 '11 at 19:00
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.