# Thread: Finding the real solutions of a logarithm

1. ## 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)

2. ## Re: Finding the real solutions of a logarithm

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)

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

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3. ## 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.

5. ## Re: Finding the real solutions of a logarithm

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

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6. ## Re: Finding the real solutions of a logarithm

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)
Originally Posted by zzephod
You are assuming that an unqualified $\displaystyle \log$ denotes $\displaystyle \log_{10}$ 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.