Finding the real solutions of a logarithm

• Aug 7th 2013, 11:33 PM
Coshy
Finding the real solutions of a logarithm
Hi guys,

What are the real solutions, in any, of log(x) = 2log(3) + log(5) - log(2) - 1
(all logs are to base 10)

• Aug 7th 2013, 11:59 PM
zzephod
Re: Finding the real solutions of a logarithm
Quote:

Originally Posted by Coshy
Hi guys,

What are the real solutions, in any, of log(x) = 2log(3) + log(5) - log(2) - 1
(all logs are to base 10)

\begin{aligned} \log(x) &= 2\log(3) + \log(5) - \log(2) -1 \\ &= \log(3^2) + \log(5) -\log(2) -1 \\ &= \log( (3^3\times 5)/(2 \times b)) \end{aligned}

where $b$ is the base for the logarithms, so $x= ...$

.
• Aug 8th 2013, 05:36 AM
HallsofIvy
Re: Finding the real solutions of a logarithm
IF you know the "laws of logarithms": a log(b)= log(b^a) and log(a)+ log(b)= log(ab), this problem is easy. If you do NOT the you should not be attempting a problem like this.
• Aug 8th 2013, 07:28 PM
ibdutt
Re: Finding the real solutions of a logarithm
• Aug 9th 2013, 07:51 AM
zzephod
Re: Finding the real solutions of a logarithm
Quote:

Originally Posted by ibdutt

You are assuming that an unqualified $\log$ denotes $\log_{10}$ which might be the case in high school maths classes but is not in the wider world.

.
• Aug 9th 2013, 08:36 AM
Plato
Re: Finding the real solutions of a logarithm
Quote:

Originally Posted by Coshy
What are the real solutions, in any, of log(x) = 2log(3) + log(5) - log(2) - 1
(all logs are to base 10)

Quote:

Originally Posted by zzephod
You are assuming that an unqualified $\log$ denotes $\log_{10}$ which might be the case in high school maths classes but is not in the wider world..

The OP clearly states they are indeed base 10.