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Mathematics 14 Online
OpenStudy (anonymous):

solve for x: 25^x+1*125^x=25^3x-2. Please show me the steps

OpenStudy (anonymous):

hey please clarify the equation with use of brackets or either use the equation tool here

OpenStudy (anonymous):

25^(x+1)*125^(x)=25^(3x-2)

OpenStudy (lalaly):

\[\large{(5^2)^{(x+1)}\times (5^3)^x = (5^2)^{(3x-2)}}\]

OpenStudy (lalaly):

\[\large{5^{2x+2} \times 5^{3x}= 5^{6x-4}}\]

OpenStudy (lalaly):

\[\large{5^{2x+2+3x} = 5^{6x-4}}\]

OpenStudy (lalaly):

2x+2+3x=6x-4 5x+2=6x-4 6=x

OpenStudy (anonymous):

the answer is 6. Please can you try again

OpenStudy (lalaly):

yeah i got it hehe

OpenStudy (anonymous):

Thanks

OpenStudy (lalaly):

your welcome =D

OpenStudy (lalaly):

SASO=p

OpenStudy (anonymous):

There are two ways to look at this problem: 1) Take a log base 10 on both side of the equation and you end up with a linear equation. this is how it goes: (2x+ 2)*log5 + 3x*log5 = (6x-4)*log5 => (2x+2+3x) *log5 = 6x-4 => 5x+2=6x-4 solve for x 2) find the common base as lalaly just solve above

OpenStudy (anonymous):

Thanks

OpenStudy (anonymous):

you are welcome

OpenStudy (lalaly):

LOL i dont have conversations on ppls posts ... talk to me somewhere esle =D

OpenStudy (sasogeek):

i was just about to say that lol :P i was compelled to. deleting the post now

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