Original post by skittles100
how would you solve this equation for x?
2^2x+3 = 3^3x+2
2^2x+3 = 3^3x+2
See the attached
From there you can do it I guess?
LOL SILLY MISTAKE
log that 2 too
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(edited 8 years ago)
Original post by C0balt
See the attached
From there you can do it I guess?
LOL SILLY MISTAKE
log that 2 too
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From there you can do it I guess?
LOL SILLY MISTAKE
log that 2 too
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attatched? why would i log the 2?
Original post by skittles100
attatched? why would i log the 2?
Ok I ****ed up here it is
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Original post by C0balt
Original post by skittles100
thanks! how would you solve this?
what can one be rewritten as?
(edited 8 years ago)
Original post by whydoidothis?
what can one be rewritten as?
2log5(x) can be written as log5(x^2)
then log5((x^2)/(3x))=1
then x^2/3x=5^1
x^2/3x=5
And well, I forgot what you do next...
Original post by Dieselblue
2log5(x) can be written as log5(x^2)
then log5((x^2)/(3x))=1
then x^2/3x=5^1
x^2/3x=5
And well, I forgot what you do next...
then log5((x^2)/(3x))=1
then x^2/3x=5^1
x^2/3x=5
And well, I forgot what you do next...
I think he means 1= log5(5)
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Original post by C0balt
yeah thats what i meant, did you manage to do it OP?
Original post by skittles100
how would you solve this equation for x?
2^2x+3 = 3^3x+2
2^2x+3 = 3^3x+2
Just out of interest, do you mean:
2^(2x+3) = 3^(3x+2)
or
(2^2x)+3 = (3^3x)+2
Original post by AG98
Just out of interest, do you mean:
2^(2x+3) = 3^(3x+2)
or
(2^2x)+3 = (3^3x)+2
2^(2x+3) = 3^(3x+2)
or
(2^2x)+3 = (3^3x)+2
I mean the first one, and do these type of questions come up in a c2 exam generally?
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Original post by skittles100
I mean the first one, and do these type of questions come up in a c2 exam generally?
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If it's the first one my photo was wrong. This one is right
Also there's always one log question which is quite hard
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Are you guys just writing your logs weirdly or do you not have bases?
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Original post by C0balt
If it's the first one my photo was wrong. This one is right
Also there's always one log question which is quite hard
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Also there's always one log question which is quite hard
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Thank you! I was struggling on this for quite a while, now it makes sense
Original post by Isis_on_the_cake
I am lazy
You are allowed to miss bases out for base 10 I think
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Original post by Isis_on_the_cake
When you write something like log6,
the base is automatically assumed to be 10.
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