# Question #d4055

Mar 6, 2016

You need to clarify the question
To format use: hash symbol log_5(2x+1) hash symbol
This looks like: $L o {g}_{5} \left(2 x + 1\right)$

#### Explanation:

Assumption: The question is:

${\log}_{3} \left(x\right) = - 2$ ......................................(1)
${\log}_{5} \left(2 x + 1\right) - {\log}_{5} \left(x\right) = 2$................(2)
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$\textcolor{b l u e}{\text{Pre amble:}}$

Compare to $\text{ } {\log}_{10} \left(x\right) = 3$
Another way of writing this is: ${10}^{3} = x$

Also ${\log}_{z} \left(a\right) - {\log}_{z} \left(b\right) = {\log}_{z} \left(\frac{a}{b}\right)$

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$\textcolor{b l u e}{\text{Consider equation (1)}}$

Write as $\text{ } {3}^{- 2} = x$

$\textcolor{b l u e}{\implies x - \frac{1}{3} ^ 2 = \frac{1}{9}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left({1}_{a}\right)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Consider equation (2)}}$

Write as:$\text{ } {\log}_{5} \left(\frac{2 x + 1}{2}\right) = 2$

$\implies {5}^{2} = \frac{2 x + 1}{2}$

$\implies 25 = x + \frac{1}{2}$

$\textcolor{b l u e}{\implies x = 24.5} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left({2}_{a}\right)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{red}{\text{Either I have interpreted the questions incorrectly or}}$
$\textcolor{red}{\text{the two equation are not connected!}}$
$\textcolor{red}{\text{It could be that the question is wrong!}}$

$\textcolor{m a \ge n t a}{\text{Equation "(2_a) !=" Equation "(1_a)" "->" contradiction}}$

Mar 6, 2016

A different interpretation of question

#### Explanation:

Assumption:

$\log \left({3}^{x}\right) = - 2$................................(1)
$\log \left({5}^{2 x + 1}\right) - \log \left({5}^{x}\right) = 2$.... ..(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Consider equation (1)

Write as $\text{ } x \log \left(3\right) = 2$

$\textcolor{b l u e}{\implies x = \frac{2}{\log} \left(3\right) \approx 4.1918}$ to 4 decimal places

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider equation (2)

Write as $\text{ } \left(2 x - 1\right) \log \left(5\right) - x L o g \left(5\right) = 2$

$\implies \log \left(5\right) \left(2 x - 1 - x\right) = 2$

$x - 1 = \frac{2}{\log} \left(5\right)$

$\textcolor{b l u e}{x = \frac{2}{\log} \left(5\right) + 1 \approx 3.8613}$ to 4 decimal places