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**tony123****Member**- Registered: 2007-08-03
- Posts: 216

Find the real solutions of the two equations

(1)

(2)

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,355

hi tony123,

I think these are valid:

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**tony123****Member**- Registered: 2007-08-03
- Posts: 216

There are some mistakes in your solution dear Bob

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,355

With one or both?

Please give more details.

LATER EDIT:

Ok I see what is wrong (I think) with each one, so it's back to the drawing board for me. This may take a while.

Bob

*Last edited by Bob (2022-01-12 01:41:15)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,355

Hurray! I've got an answer for (2) that is confirmed by wolfram alpha. Very busy today so I'll post it when I've got more time.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,355

Here we go:

All real log functions are undefined for x ≤ 0 so x > 11 or the log powers don't work.

Replace x by X = x - 11

The equation becomes

For brevity I will use all logs in base 2 from now on.

case 1. log(X) = 0

case 2. if log(X) ≠ 0 then it can be cancelled leaving

Raise 2 to the power of each side (you could also call this antilog}

The quadratic formula has one negative answer which contradicts the domain for X so

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,355

And now for Q1:

First some preliminaries:

If

... Aand if

... BAnd now the question

All logs in base 7

using A becomes

Divide by x

Replace 1 by log(7) and simplify the logs

Using A again gives

I will replace log(x) with X

Now what does the graph of the left hand side function look like?

It has two components. The first is a standard power graph, going through (0,1) and rising thereafter. The second is is similar but, as 3/7 is under 1, it will come down from large y values to the left of the y axis, again go through (0,1) and then drop off to zero as X gets larger.

Together the second component will dominate when X is negative; then the first will dominate. So the graph will start high, drop down to (0,2) and at some stage rise again. Note that (0,2) gives one solution to the question.

Differentiating using B

The ln(3/7) is negative so this will give one turning point. It must be a minimum because of the overall shape already established; (0,2) could be the minimum; otherwise there will be exactly one more solution.

I can see that X = 1 gives x = 7 leading to 11/7 + 3/7 = 14/7 = 2. So here is the second solution.

So the equation has two solutions, x = 1 and x = 7.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**tony123****Member**- Registered: 2007-08-03
- Posts: 216

Now we can say great job

Thanks bob

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