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OpenStudy (piercetheveil47):
2(3)^x = 3^x + 1?
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OpenStudy (bibby):
\(\large 2*3^x=3^{x+1}\)? what are we doing? solving for x?
OpenStudy (piercetheveil47):
yes @bibby
OpenStudy (bibby):
divide both sides by 3^x
OpenStudy (phi):
are you sure the problem is not
\[ 2\cdot 3^x = 3^x + 1 \]
OpenStudy (bibby):
i figured it was a log problem
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OpenStudy (phi):
there is no solution the other way
OpenStudy (bibby):
oh yeah, my bad
OpenStudy (bibby):
2=3
OpenStudy (piercetheveil47):
@phi yes sorry! i just looked and you're correct, i was mistaken
OpenStudy (phi):
yes and 2=3 means no solution (most experts agree that 2 is not equal to 3)
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OpenStudy (phi):
but to solve, you can still use bibby's advice
or you can "combine like terms"
OpenStudy (piercetheveil47):
ohhhh! i get it. so the answer is x = no solution?
OpenStudy (phi):
if you temporarily rename 3^x as y, you have
2 y = y + 1
OpenStudy (phi):
now solve for y. what do you get ?
OpenStudy (phi):
***ohhhh! i get it. so the answer is x = no solution?***
only if the problem is
\[ 2\cdot 3^x = 3^{x+1} \]
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OpenStudy (phi):
to solve
\[ 2\cdot 3^x = 3^x + 1 \]
temporarily rename 3^x as y
\[ 2y= y+1 \]
solve for y by adding -y to both sides.
can you do that ?
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