June 26th, 2018, 02:24 PM  #1 
Newbie Joined: Apr 2018 From: Banovo Brdo Posts: 12 Thanks: 0  Logarithms
$\left(\sqrt{\log_7\left(2x1\right)}\log_7\sqrt{4x^2+4x+1}1\right)\cdot \log_7\leftx+7\right=3\log_7\sqrt[3]{x+7}$ What are the solutions to this equation? I got 6 and 4, but I think it should be 6 and 5? Thanks. EDIT: Never mind, solved. Last edited by greg1313; June 26th, 2018 at 06:51 PM. 
June 26th, 2018, 02:48 PM  #2 
Senior Member Joined: Aug 2012 Posts: 2,157 Thanks: 631 
Hi bongcloud. It's legal where I live.

June 26th, 2018, 03:54 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,285 Thanks: 1967 
I'll assume that $x + 7 > 0$. If $7^{n^2} = 2x  1$, where $n \geqslant 0$ and $2x  1 > 0$, $n  n^2  1 = 1$, and so $n$ is 1 or 0. Hence $2x  1$ = 7 or 1, and so $x$ = 4 or 1. 

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