# logarithmic equation

• Oct 21st 2012, 03:30 AM
skaters
logarithmic equation
• Oct 21st 2012, 03:39 AM
Prove It
Re: logarithmic equation
\displaystyle \begin{align*} \frac{1}{2}\log{(5 + 7x)} - \log{9} + \log{2} &= 1 - \frac{1}{2}\log{(7 + 2x)} \\ \frac{1}{2}\log{(5 + 7x)} + \frac{1}{2}\log{(7 + 2x)} + \log{\frac{2}{9}} &= 1\\ \frac{1}{2}\log{\left[ (5 + 7x)(7 + 2x) \right]} + \log{\frac{2}{9}} &= 1 \\ \log{\left\{ \left[ (5 + 7x)(7 + 2x) \right]^{\frac{1}{2}} \right\}} + \log{\frac{2}{9}} &= 1 \\ \log{\left\{ \frac{2}{9}\left[ (5 + 7x)(7 + 2x) \right]^{\frac{1}{2}} \right\}} &= 1 \\ \frac{2}{9}\left[ (5 + 7x)(7 + 2x) \right]^{\frac{1}{2}} &= e^1 \\ \left[ (5 + 7x)(7 + 2x) \right]^{\frac{1}{2}} &= \frac{9e}{2} \\ (5 + 7x)(7 + 2x) &= \frac{81e^2}{4} \\ 35 + 10x + 49x + 14x^2 &= \frac{81e^2}{4} \\ 14x^2 + 59x + 35 + \frac{81e^2}{4} &= 0 \end{align*}

You can now solve this using the Quadratic Formula.
• Oct 21st 2012, 04:18 AM
skaters
Re: logarithmic equation
Can you do quadratic aswell - I havent done any with e2
• Oct 21st 2012, 04:26 AM
Prove It
Re: logarithmic equation
Quote:

Originally Posted by skaters
Can you do quadratic aswell - I havent done any with e2

Yes I can, but I won't. It works exactly the same way, your c value just happens to be \displaystyle \begin{align*} 35 + \frac{81e^2}{4} \end{align*}.
• Oct 21st 2012, 09:39 AM
Soroban
Re: logarithmic equation
Hello, skaters!

I assume that those logs are all base-ten.

Quote:

$\tfrac{1}{2}\log{(5 + 7x)} - \log{9} + \log{2} \:=\: 1 - \tfrac{1}{2}\log{(7 + 2x)}$

$\begin{array}{cc}\text{We have:} & \tfrac{1}{2}\log{(7x+5)} + \tfrac{1}{2}\log{(2x+7)} + \log{\tfrac{2}{9}} \:=\: 1 \\ \\ & \log(7x+5)^{\frac{1}{2}} + \log(2x +7)^{\frac{1}{2}} + \log(\tfrac{2}{9}) \:=\:1 \\ \\ \text{Combine:} & \log\left[\tfrac{2}{9}(7x+5)^{\frac{1}{2}}(2x+7)^{\frac{1}{2 }}\right] \:=\:\log(10) \\ \\ \text{"Un-log":} & \tfrac{2}{9}(7x+5)^{\frac{1}{2}}(2x+7)^{\frac{1}{2 }} \;=\;10 \\ \\ & (7x+5)^{\frac{1}{2}}(2x+7)^{\frac{1}{2}} \;=\;45 \\ \\ \text{Square:} & (7x + 5)(2x+7) \;=\;2025 \\ \\ & 14x^2 + 59x + 25 \;=\;2025 \\ \\ & 14x^2 + 59x - 1990 \;=\;0 \\ \\ \text{Factor:} & (x-10)(14x-199) \;=\;0 \\ \\ \text{Hence:} & x \;=\;10,\,\text{-}\frac{199}{14} \end{array}$

We must discard the negative root.

Therefore: . $x\,=\,10$
• Oct 22nd 2012, 11:40 AM
skaters
Re: logarithmic equation
Thanks ! :)