exponential equality: 2^x + 2^(x+1)= 4^x + 4^(x+1)

[smsmith]

I am a bit rusty. I solved this by elimination. However, no matter what I tried I could not seem to figure out an algebraic solution. Any input would be appreciated.

2^x + 2^(x+1)= 4^x + 4^(x+1)

Thanks in advance,
S M Smith

 

[tkhunny]

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?

 

[galactus]

Take log of both sides:

\(\displaystyle \L\\ln(2^{x}+2^{x+1})=ln(4^{x}+4^{x+1})\)

=\(\displaystyle \L\\xln(2)+ln(3)=2xln(2)+ln(5)\)

=\(\displaystyle \L\\xln(2)-2xln(2)=ln(5)-ln(3)\)

=\(\displaystyle \L\\x(ln(2)-2ln(2))=ln(5)-ln(3)\)

=\(\displaystyle \L\\x=\frac{ln(5)-ln(3)}{ln(2)-2ln(2)}\)

=\(\displaystyle \H\\\frac{ln(\frac{3}{5})}{ln(2)}\)

 

[soroban]

Hello, smsmith!

$\displaystyle 2^x\,+\,2^{x+1}\;= \;4^x\, +\, 4^{x+1}$

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: $\displaystyle \:\ln\left(2^x\right) \;= \;\ln\left(\frac{3}{5}\right)\;\;\Rightarrow\;\;x\cdot\ln2\;=\;\ln\left(\frac{3}{5}\right)$

\(\displaystyle \text{Therefore: }\L\:x\;=\;\frac{\ln\left(\frac{3}{5}\right)}{\ln 2}\)