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April 22nd, 2014, 06:20 PM  #1 
Newbie Joined: Apr 2014 From: South Florida Posts: 8 Thanks: 1  Exponential Quadratic
This problem popped up on my homework: $\displaystyle 7^{2x+1}15•7^x+2=0$ Is the $\displaystyle 2x+1$ just a typo? Can this be done? Note that my teacher is very prone to making typos. I did the problem assuming that it was just $\displaystyle 7^{2x}$ and got $\displaystyle x=log_7\left({15\pm\sqrt{218}\over 2}\right)$. Thanks. 
April 22nd, 2014, 06:30 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
\begin{align*} 7^{2x+1}15 \cdot 7^x+2 &= 0 \\ 7 \cdot 7^{2x}15 \cdot 7^x+2 &= 0 \\ \left( 7 \cdot 7^x  1\right) \left( 7^x  2 \right) &= 0 \\ \left( 7^{x+1}  1\right) \left( 7^x  2 \right) &= 0 \\ \end{align*} So $$\left(x + 1\right)\log{7} = 0 \; \Longrightarrow \; x = 1$$ or $$x\log{7} = \log{2} \; \Longrightarrow \; x = \frac{\log{2}}{\log{7}}$$ I like my numbers better than yours. Last edited by v8archie; April 22nd, 2014 at 06:36 PM. 
April 22nd, 2014, 06:53 PM  #3  
Newbie Joined: Apr 2014 From: South Florida Posts: 8 Thanks: 1  Quote:
 
April 22nd, 2014, 08:38 PM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
An alternate approach: let y = 7^x and replace 7^(2x+1) with 7y^2.


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exponential, quadratic 
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