Logarithm equation help

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SteveMcs
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#1
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#1
Having some trouble with one question to do with logs.
(note, logs are all to the base 10)

Solve for x:

log(19x) = 3.9 - 2.1x

I just don't know how to tackle it with the "non-logged" x on the RHS

if it was just logA = B i would say A = 10^B but that doesn't help me find what x is.

I'm not asking anyone to do the question for me, but a step in the right direction would be appreciated.
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TenOfThem
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#2
Report 7 years ago
#2
(Original post by SteveMcs)
Having some trouble with one question to do with logs.
(note, logs are all to the base 10)

Solve for x:

log(19x) = 3.9 - 2.1x

I just don't know how to tackle it with the "non-logged" x on the RHS

if it was just logA = B i would say A = 10^B but that doesn't help me find what x is.

I'm not asking anyone to do the question for me, but a step in the right direction would be appreciated.
Are you sure that this is the question as it would be solved using numerical rather than algebraic methods
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SteveMcs
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#3
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(Original post by TenOfThem)
Are you sure that this is the question as it would be solved using numerical rather than algebraic methods
100% sure. Only way i could think was by graphing both sides and giving an approximation but they want the answer to 2 dp. I don't think i'd be able to give an approximation that accurate.
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TenOfThem
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#4
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(Original post by SteveMcs)
100% sure. Only way i could think was by graphing both sides and giving an approximation but they want the answer to 2 dp. I don't think i'd be able to give an approximation that accurate.
What course are you doing?
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SteveMcs
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#5
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#5
This isn't actually for my course. I'm doing BSc Actuarial Science. But this is for my partner who's doing a Numerical Skills for Scientists module in Environmental Earth Science
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SteveMcs
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#6
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#6
I got the answer in this example to be 1.21 to 2dp as there is a sign change when subbing x=1.205 and x=1.21. There's got to be an easier way of solving this?
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davros
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#7
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#7
(Original post by SteveMcs)
I got the answer in this example to be 1.21 to 2dp as there is a sign change when subbing x=1.205 and x=1.21. There's got to be an easier way of solving this?
There isn't an exact method for solving an equation like this so some form of numerical approximation or iterative process is going to be necessary. Has your partner been taught specific methods that are to be used in cases like this, because it's almost certainly going to be just a case of choosing the right numerical method
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Mr M
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#8
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#8
(Original post by SteveMcs)
I got the answer in this example to be 1.21 to 2dp as there is a sign change when subbing x=1.205 and x=1.21. There's got to be an easier way of solving this?
The easiest way is to use iteration. You need to rearrange the equation into the form x_{n+1}=f(x_n).

In the case a suitable rearrangement is x_{n+1}=\frac{3.9-\log_{10} (19x_n)}{2.1}

Type 1 = into your calculator.

Now type \frac{3.9-\log (19 ANS)}{2.1} and press = a few times.
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