I don't exactly know where to begin with this one. solve: 5^2x+1=1/27
E ekrutsch New member Joined Nov 30, 2009 Messages 1 Nov 30, 2009 #1 I don't exactly know where to begin with this one. solve: 5^2x+1=1/27
D Deleted member 4993 Guest Nov 30, 2009 #2 ekrutsch said: I don't exactly know where to begin with this one. solve: 5^2x+1=1/27 Click to expand... Is it: \(\displaystyle 5^{2x+1}\) or \(\displaystyle 5^{2x} + 1\) or \(\displaystyle 5^2x + 1\) Also re-check all the numbers for accuracy.
ekrutsch said: I don't exactly know where to begin with this one. solve: 5^2x+1=1/27 Click to expand... Is it: \(\displaystyle 5^{2x+1}\) or \(\displaystyle 5^{2x} + 1\) or \(\displaystyle 5^2x + 1\) Also re-check all the numbers for accuracy.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Nov 30, 2009 #3 Hello, ekrutsch! \(\displaystyle \text{Solve: }\;5^{2x+1} \:=\:\frac{1}{27}\) Click to expand... \(\displaystyle \text{Take logs: }\;\ln\left(5^{2x+1}\right) \;=\;\ln\left(\frac{1}{27}\right) \;=\;\ln\left(\frac{1}{3^3}\right) \;=\;\ln\left(3^{-3}) \;=\;-3\ln(3)\) \(\displaystyle \text{We have: }\;(2x+1)\ln(5) \;=\;-3\ln(3) \quad\Rightarrow\quad 2x + 1 \;=\;\frac{-3\ln(3)}{\ln(5)}\) . . . . . . \(\displaystyle 2x \;=\;-\frac{3\ln(3)}{\ln(5)} - 1 \quad\Rightarrow\quad x \;=\;-\frac{1}{2}\bigg[\frac{3\ln(3)}{\ln(5)} + 1\bigg] \;\approx\;-1.524\)
Hello, ekrutsch! \(\displaystyle \text{Solve: }\;5^{2x+1} \:=\:\frac{1}{27}\) Click to expand... \(\displaystyle \text{Take logs: }\;\ln\left(5^{2x+1}\right) \;=\;\ln\left(\frac{1}{27}\right) \;=\;\ln\left(\frac{1}{3^3}\right) \;=\;\ln\left(3^{-3}) \;=\;-3\ln(3)\) \(\displaystyle \text{We have: }\;(2x+1)\ln(5) \;=\;-3\ln(3) \quad\Rightarrow\quad 2x + 1 \;=\;\frac{-3\ln(3)}{\ln(5)}\) . . . . . . \(\displaystyle 2x \;=\;-\frac{3\ln(3)}{\ln(5)} - 1 \quad\Rightarrow\quad x \;=\;-\frac{1}{2}\bigg[\frac{3\ln(3)}{\ln(5)} + 1\bigg] \;\approx\;-1.524\)
