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### Solving a Logarithm Equation by Substitution

```Date: 12/18/2007 at 15:37:54
From: Eugene
Subject: Evaluating "x" for a log equation

I have a "thinking" question which states:

solve (x+1)^(log(x+1)) = 100(x+1)

I recently learned logs and I find it difficult to apply it to solving
equations.  So far I divided both sides by (x+1).  Therefore:

(x+1)^(log(x+1))/(x+1) = 100

then

(x+1)^(log(x+1)-1) = 100

then i applied log to the entire equation

log[(x+1)^(log(x+1)-1)] = log100

then the power comes in front

(log(x+1)-1)*log(x+1) = log100

this is the part where I'm not sure if I'm doing it correctly.  I
divided the entire equation by 1og therefore i get

x*(x+1) = 100
x^2 + x = 100
x^2 + x - 100 = 0

I applied the quadratic formula and got 9.51249 and -10.51249

I tried to put these back into the original equation, but they don't
agree, so clearly I went wrong somewhere, but I can't figure out where.

```

```
Date: 12/18/2007 at 17:03:13
From: Doctor Peterson
Subject: Re: Evaluating

Hi, Eugene.

You wrote:

>then the power comes in front
>
>  (log(x+1)-1)*log(x+1) = log100
>
>this is the part where I'm not sure if I'm doing it correctly.  I
>divided the entire equation by 1og

No!  No!  No!

;-)

"Log" is not a number, so you can't divide by it.  Even if it had
been a number, you can't divide every factor by something!

What you CAN do here is to notice that "log(x+1)" occurs twice, and
in fact that is the only form in which x appears.  So it can simplify
things if you give that a name, say

y = log(x+1)

Now put that name in place of the expression:

(y - 1) * y = log(100)

That ugly log on the right isn't so nice--until you realize that this
is a base ten log, and you can evaluate it easily.  Now solve for y,
and then you can find what x is from there.

If you're new to logs, this is quite a tricky problem, and you did
very well as far as you got.  And doing that check at the end is
great--too many students don't bother to check, so they never realize
that their solution was nonsense--and so they never learn.  Now you
will.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 12/18/2007 at 17:43:37
From: Eugene
Subject: Evaluating "x" for a log equation

AH!  Thanks a lot.  I was confused into dividing the log out because
in one example we did in class we had a scenario where:

log[base 3]((5x+10/(x-1)) = log[base 3](3x+2)

and the log[base 3] was divided to produce

(5x+10)/(x-1) = (3x+2)

The above is correct, right?...  (I hope it is.....)

Thanks again.

```

```
Date: 12/18/2007 at 22:45:36
From: Doctor Peterson
Subject: Re: Evaluating

Hi, Eugene.

Yes, that's correct.

I almost took the time to point out the difference with this kind of
step in my first reply, but decided to hold that for later.  Thanks

What you're doing here is like the sort of "canceling" you do by
dividing both sides of an equation by the same thing; but it depends
on a different fact.  Just as you "undo" a multiplication by
performing the inverse operation, division, here you are undoing the
logs by doing the inverse operation, namely an exponential function.
Canceling the logs is done by raising the base, 3, to the quantities
on each side of the equation, and using the fact that 3^log_3(a) = a.
As a result, if log(a) = log(b), then a = b.  (This can also be
expressed by saying that the log is a one-to-one function; this is
equivalent to the fact that it has an inverse.)

You can only undo an operation or function in this way when it is the
last one performed (following the order of operations), so that it is
"exposed" on the "outside" of the expression.  The reason you couldn't
do it in your problem is that the logs were not "on the outside"; the
last operation on the left side was a multiplication, so raising 10 to
that power would not have the desired effect.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Logs

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