youngmath
solve without using calculator
5ln(x-2)-2=28
help please, thank you
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Champs
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5ln(x-2) = 30
Champs
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ln(x-2) = 6
Champs
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ln or log ?
Mimi_x3
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\( 5ln(x-2) = 30 \)
\(ln(x-2) = 6\)
\(log_e(x-2) = 6\)
\((x-2) = e^{6}\)
Are you able to do it now?
youngmath
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ln
youngmath
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im stuck at the (x-2)=e^6
Mimi_x3
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bring the \(2\) to the RHS
youngmath
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i know the answer is e^6 +2 but i dont know where the =2 comes from
youngmath
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+2
Mimi_x3
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Because when you bring the \(-2\) to the RHS it's a positive
Mimi_x3
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I don't know the correct way to say that lol
youngmath
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alright can you walk me through the part after ln(x-2)=6
Mimi_x3
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Well you know that \(ln\) can also be written as \(log_e\) right?
youngmath
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yea
youngmath
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so you get e^6= (x-2) right
Mimi_x3
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yeah \((x-2) = e^{6}\)
youngmath
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so what do you do next
Mimi_x3
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well i just told you before :)
youngmath
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bring the 2 to the RHS?
youngmath
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ooooohhhhhhhh
xD
duh x = e^6 + 2
Mimi_x3
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yes; you dont really bring it was called something else which forgot. just simply drag it to RHS. But there is a better name for it lol
youngmath
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i forgot i was solving for x , silly me
youngmath
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i more last question what do you do for
e^x-3 = 16
Mimi_x3
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\[ e^x - 3 =16\]
What have you tried?
youngmath
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e^x-3 + 4 =20
Mimi_x3
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Um, that is not right..
Mimi_x3
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drag the \(-3) to the rhs
youngmath
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but its an exponent
x-3 is the exp. for e
Mimi_x3
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\[ e^x = 16+ 3\]
Mimi_x3
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\[ e^x = 19\]
Mimi_x3
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Take ln both sides
\[xlne = ln19\]
youngmath
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the answer is ln(16)+3
youngmath
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my teacger myst be wring though . she gave me that answer
youngmath
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my teacher was wron. she gave me that answer
Mimi_x3
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well there is a low chance that wolfram alpha is wrong
m_charron2
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oh, well, it looks like we have a paranthesis confusion here. when writing e^x-3 = 16, it's important that you write the parantheses aroudnthe (x-3), else we can't know that the whole thing is the exponant. So we have this right?
\[e ^{x-3} = 16\]