June 26th, 2018, 01:24 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 05:51 PM. 
June 26th, 2018, 01:48 PM  #2 
Senior Member Joined: Aug 2012 Posts: 1,960 Thanks: 547 
Hi bongcloud. It's legal where I live.

June 26th, 2018, 02:54 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,191 Thanks: 1649 
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. 

Tags 
logarithms 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
logarithms  ron246  Algebra  5  May 22nd, 2014 06:59 AM 
Logarithms  Mr Davis 97  Algebra  1  May 3rd, 2014 06:34 PM 
logarithms!  Mindless_1  Algebra  2  November 14th, 2012 12:02 PM 
Logarithms  BrianMX34  Algebra  7  September 6th, 2012 01:32 PM 
Logarithms  MissC  Algebra  3  May 26th, 2011 12:49 AM 