exponential equality: 2^x + 2^(x+1)= 4^x + 4^(x+1)
- Thread starter smsmith
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What does "elimination" mean in this context?
Some Hints:
\(\displaystyle 4 = 2^{2}\)
\(\displaystyle 2^{x+1} = 2^{x}*2^{1} = 2^{x}*2\)
Some Work:
\(\displaystyle 2^{x} + 2^{x+1} = 2^{x}(1 + 2) = 2^{x}(3)\)
\(\displaystyle 4^{x} + 4^{x+1} = 4^{x}(1 + 4) = 4^{x}(5) = 2^{2x}(5) = (2^{x})^{2}(5)\)
Thus:
\(\displaystyle 2^{x}(3) = (2^{x})^{2}(5)\)
Honest, it's getting better.
Can you go from there?
Some Hints:
\(\displaystyle 4 = 2^{2}\)
\(\displaystyle 2^{x+1} = 2^{x}*2^{1} = 2^{x}*2\)
Some Work:
\(\displaystyle 2^{x} + 2^{x+1} = 2^{x}(1 + 2) = 2^{x}(3)\)
\(\displaystyle 4^{x} + 4^{x+1} = 4^{x}(1 + 4) = 4^{x}(5) = 2^{2x}(5) = (2^{x})^{2}(5)\)
Thus:
\(\displaystyle 2^{x}(3) = (2^{x})^{2}(5)\)
Honest, it's getting better.
Can you go from there?Take log of both sides:
\(\displaystyle \L\\ln(2^{x}+2^{x+1})=ln(4^{x}+4^{x+1})\)
=\(\displaystyle \L\\xln(2)+ln(3)=2xln(2)+ln(5)\)
=\(\displaystyle \L\\xln(2)-2xln(2)=ln(5)-ln(3)\)
=\(\displaystyle \L\\x(ln(2)-2ln(2))=ln(5)-ln(3)\)
=\(\displaystyle \L\\x=\frac{ln(5)-ln(3)}{ln(2)-2ln(2)}\)
=\(\displaystyle \H\\\frac{ln(\frac{3}{5})}{ln(2)}\)
\(\displaystyle \L\\ln(2^{x}+2^{x+1})=ln(4^{x}+4^{x+1})\)
=\(\displaystyle \L\\xln(2)+ln(3)=2xln(2)+ln(5)\)
=\(\displaystyle \L\\xln(2)-2xln(2)=ln(5)-ln(3)\)
=\(\displaystyle \L\\x(ln(2)-2ln(2))=ln(5)-ln(3)\)
=\(\displaystyle \L\\x=\frac{ln(5)-ln(3)}{ln(2)-2ln(2)}\)
=\(\displaystyle \H\\\frac{ln(\frac{3}{5})}{ln(2)}\)
Hello, smsmith!
Get the same base . . .
The right side is: \(\displaystyle \,(2^2)^x\,+\,(2^2)^{x+1}\;=\;2^{2x}\,+\,2^{2x+2}\)
The equation becomes: \(\displaystyle \:2^x\,+\,2^{x+1} \;= \;2^{2x}\,+\,2^{2x+2}\)
. . and we have: \(\displaystyle \:2^{2x+2}\,+\,2^{2x}\,-\,2^{x+1}\,-\,2^x\;=\;0\)
Since \(\displaystyle 2^x\,\neq\,0,\) divide by \(\displaystyle 2^x:\;\;2^{x+2}\,+\,2^x\,-\,2^1\,-\,1 \;=\;0\)
and we have: \(\displaystyle \:2^{x+2}\,+\,2^x\,-\,3\;=\;0\)
. . . . . . . . . . \(\displaystyle 2^2\cdot2^x\,+\,2^x\,-\,3\;=\;0\)
. . . . . . . . . . .\(\displaystyle 4\cdot2^x\,+\,2^x\,-\,3\;=\;0\)
. . . . . . . . . . . . . . \(\displaystyle 5\cdot2^x\,-\,3\;=\;0\)
. . . . . . . . . . . . . . . . . . \(\displaystyle 2^x\;=\;\frac{3}{5}\)
Take logs: \(\displaystyle \:\ln\left(2^x\right) \;= \;\ln\left(\frac{3}{5}\right)\;\;\Rightarrow\;\;x\cdot\ln2\;=\;\ln\left(\frac{3}{5}\right)\)
\(\displaystyle \text{Therefore: }\L\:x\;=\;\frac{\ln\left(\frac{3}{5}\right)}{\ln 2}\)
Get the same base . . .
The right side is: \(\displaystyle \,(2^2)^x\,+\,(2^2)^{x+1}\;=\;2^{2x}\,+\,2^{2x+2}\)
The equation becomes: \(\displaystyle \:2^x\,+\,2^{x+1} \;= \;2^{2x}\,+\,2^{2x+2}\)
. . and we have: \(\displaystyle \:2^{2x+2}\,+\,2^{2x}\,-\,2^{x+1}\,-\,2^x\;=\;0\)
Since \(\displaystyle 2^x\,\neq\,0,\) divide by \(\displaystyle 2^x:\;\;2^{x+2}\,+\,2^x\,-\,2^1\,-\,1 \;=\;0\)
and we have: \(\displaystyle \:2^{x+2}\,+\,2^x\,-\,3\;=\;0\)
. . . . . . . . . . \(\displaystyle 2^2\cdot2^x\,+\,2^x\,-\,3\;=\;0\)
. . . . . . . . . . .\(\displaystyle 4\cdot2^x\,+\,2^x\,-\,3\;=\;0\)
. . . . . . . . . . . . . . \(\displaystyle 5\cdot2^x\,-\,3\;=\;0\)
. . . . . . . . . . . . . . . . . . \(\displaystyle 2^x\;=\;\frac{3}{5}\)
Take logs: \(\displaystyle \:\ln\left(2^x\right) \;= \;\ln\left(\frac{3}{5}\right)\;\;\Rightarrow\;\;x\cdot\ln2\;=\;\ln\left(\frac{3}{5}\right)\)
\(\displaystyle \text{Therefore: }\L\:x\;=\;\frac{\ln\left(\frac{3}{5}\right)}{\ln 2}\)