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Mathematics 8 Online
OpenStudy (piercetheveil47):

2(3)^x = 3^x + 1?

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

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)

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 \[ 2\cdot 3^x = 3^{x+1} \]

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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