Algebra Pre-Algebra and Basic Algebra Math Forum

 April 22nd, 2014, 07:20 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, 07:30 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 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. Thanks from scomora Last edited by v8archie; April 22nd, 2014 at 07:36 PM.
April 22nd, 2014, 07:53 PM   #3
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Joined: Apr 2014
From: South Florida

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Quote:
 Originally Posted by v8archie \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.
I like yours a lot more too! Never thought of it like that. Thanks a lot

 April 22nd, 2014, 09:38 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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