Selekt1234
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#1
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loga x^2 + ln x^3 + 6 loga x
for ln a = 3.
how do i simply this expression?
If you know how to do can you please explain it step by step
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mqb2766
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#2
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#2
(Original post by Selekt1234)
loga x^2 + ln x^3 + 6 loga x
for ln a = 3.
how do i simply this expression?
If you know how to do can you please explain it step by step
Does loga mean log base a?I
Some common log rules at
https://www.mathsisfun.com/algebra/e...ogarithms.html
Have a think about which may be useful?
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Selekt1234
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i have done it which simplified to 17/3lnx
can anyone let me know if its wrong or right?
and part 2 of the question is loga x^2 + ln x^3 + 6 loga x=17/3lnx
but it doesnt quite make sense to me as there is no solution to it with my answer (ps logax does mean log base a)
Also i have another question which is a simoultaneous equations which is
lnx+2lny=ln3
lnx+lny=1
i have done it and i got
x=3e^2
y=0.1222
can you let me know if its correct becaseu when i plug them back in the equation i get 0,997 and i dont get 1 exactly
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mqb2766
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#4
(Original post by Selekt1234)
i have done it which simplified to 17/3lnx
can anyone let me know if its wrong or right?
and part 2 of the question is loga x^2 + ln x^3 + 6 loga x=17/3lnx
but it doesnt quite make sense to me as there is no solution to it with my answer (ps logax does mean log base a)
Also i have another question which is a simoultaneous equations which is
lnx+2lny=ln3
lnx+lny=1
i have done it and i got
x=3e^2
y=0.1222
can you let me know if its correct becaseu when i plug them back in the equation i get 0,997 and i dont get 1 exactly
Can you upload your working in both cases pls?
Last edited by mqb2766; 4 weeks ago
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boulderingislife
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#5
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#5
(Original post by Selekt1234)
how do i simply this expression?
If you know how to do can you please explain it step by step
loga x^2 + ln x^3 + 6 loga x
for ln a = 3
ln(a.x^2)/ln10 +ln(x^3) + ln(a.x)/ln10 {assuming log is to base 1}

then you get:
(Ln a +lnx^2)/ln10 +(lna+ ln x)/ln10 + lnx^3
3/ln10 +2lnx/ln10 +3/ln10 + lnx/ln10 +3lnx
6/ln10+3lnx/ln10 +3lnx
(6+3lnx)/ln10 +3lnx
(6x^3 +(3x^3)lnx)/ln10

might be wrong. Too sleepy atm!
Last edited by boulderingislife; 4 weeks ago
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davros
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#6
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#6
(Original post by Selekt1234)
i have done it which simplified to 17/3lnx
can anyone let me know if its wrong or right?
and part 2 of the question is loga x^2 + ln x^3 + 6 loga x=17/3lnx
but it doesnt quite make sense to me as there is no solution to it with my answer (ps logax does mean log base a)
Also i have another question which is a simoultaneous equations which is
lnx+2lny=ln3
lnx+lny=1
i have done it and i got
x=3e^2
y=0.1222
can you let me know if its correct becaseu when i plug them back in the equation i get 0,997 and i dont get 1 exactly
That's not correct and I would advise leaving your answer for y (when you find it!) in terms of e rather than a decimal - it makes it easier to plug back into the original equations and check your answer.
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Selekt1234
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#7
Report Thread starter 4 weeks ago
#7
is x not 3e^2
if not what is it
and according to you y would be y=e^1-ln(3e^2)
and also regarding 17/3lnx
my working out
ln(a)=3
so converted everything into Ln using change of base formula
and got 1/3lnx^2+lnx^3+lnx^2 and then turned them into all lnx
2/3lnx+3lnx+2lnx=17/3lnx
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mqb2766
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17/3 ln(x) looks good to me.
Can you upload a pic of the original question?
Last edited by mqb2766; 4 weeks ago
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davros
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#9
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#9
(Original post by Selekt1234)
is x not 3e^2
if not what is it
and according to you y would be y=e^1-ln(3e^2)
I don't get either of those values.

Take a step back and work from the basics.

You have lnx + 2lny = ln3 which means lnx + ln(y^2)  = ln3 using one of the log rules, or ln(xy^2) = ln3 applying another log rule. So what can you say about xy^2 ?

Now trying following the same principle with the second equation and see what you can deduce about xy.
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Selekt1234
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#10
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#10
x=e^1
y=1
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davros
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#11
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(Original post by Selekt1234)
x=e^1
y=1
That pair of values satisfies the 2nd of the equations: lnx + lny = 1.

Does it satisfy the 1st equation: lnx + 2lny = ln3 ?
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Selekt1234
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#12
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#12
y=e^ln3-1
x=e^2-ln3
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davros
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#13
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(Original post by Selekt1234)
y=e^ln3-1
x=e^2-ln3
If you mean this:

y = e^{ln3 - 1}

and

x = e^{2 - ln3}

then I think those are technically correct, however, you haven't simplified them as much as you can!

How can you write those answers in simplified form?
(Hint: there shouldn't be any logarithms left in the final answer.)
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Selekt1234
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#14
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#14
Also I have another question
Simplify
log3x+log9x
3 and 9=base
So I changed the base to 3
I got log3(x)/log3(9) which equals to 1/2log3(x) right then if I take the power up it will be log3(x^1/2)
So then it’s multiplication of log3(x X x^1/2) which =log3x^3\2
But when I keep the 1/2 and not put it as a power 1/2logx^2 so which would be the right answer
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davros
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#15
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#15
(Original post by Selekt1234)
Also I have another question
Simplify
log3x+log9x
3 and 9=base
So I changed the base to 3
I got log3(x)/log3(9) which equals to 1/2log3(x) right then if I take the power up it will be log3(x^1/2)
So then it’s multiplication of log3(x X x^1/2) which =log3x^3\2
But when I keep the 1/2 and not put it as a power 1/2logx^2 so which would be the right answer
You seem to have forgotten to add the other term when you converted log_9(x) to log_3(x)

If w = log_3(x)

then you end up with

w + (1/2)w  = (3w/2) = log_3(x^{3/2})
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Selekt1234
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(Original post by davros)
You seem to have forgotten to add the other term when you converted log_9(x) to log_3(x)

If w = log_3(x)

then you end up with

w + (1/2)w  = (3w/2) = log_3(x^{3/2})
initially I got what you got but however I put it on Photomath it gives me 1/2logx^2
Last edited by Selekt1234; 4 weeks ago
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davros
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#17
(Original post by Selekt1234)
initially I got what you got but however I put it on Photomath it gives me 1/2logx^2
I'm not familiar with Photomath but maybe it misunderstood what you'd entered?

Anyway, working out the algebra according to the log rules is the safe way to go
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Selekt1234
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#18
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How would I simplify
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DFranklin
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#19
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https://www.khanacademy.org/math/alg...ase-rule-intro
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