# Q1 - Equation with fractional and logarithmic exponents

1. Aug 23, 2010

### ExamFever

1. The problem statement, all variables and given/known data

4 1/(x+1) · 8 1/(x+2) = 9log3 2

2. Relevant equations

n/a - same as 1

3. The attempt at a solution

To get rid of the exponents and be able to solve for x, I first took the logarithm with base 4 of all terms:

log44 1/(x+1) · log48 1/(x+2) = log49 log3 2

This equals:

1/(x+1) · 1/(x+2) · log48 = log32 · log49

When adding the fractions the left side equals:

log48/((x+1)(x+2))

Here I already feel I am on the wrong track, but it is still possible to get rid of the fraction by multiplying by the LCM (x+1)(x+2):

log4 8 = (x+1)(x+2)log32 · log49

Now it is possible to get the x's on one side by dividing both sides by log32 · log49:

log48/(log32 · log49) = (x+1)(x+2)

The x's can be worked free as: x2+3x+2, thus:

log48/(log32 · log49) -2 = x2+3x

The right side can now be made to look like: x(x+3), however, this does not solve x!

Now I am at a loss here, I have the feeling the three exponents at the start should be freed much sooner and in a cleaner way then I tried above, but this is the closest I came to a solution for x. Some help would be appreciated!

2. Aug 23, 2010

### eumyang

You can simplify the right side before you even take the log base 4 of both sides. Note that
$$b^{\log_b n} = n$$, so

\begin{aligned} 9^{\log_3 2} &= (3^2)^{\log_3 2} \\ &= 3^{2 \log_3 2} \\ &= 3^{\log_3 4} \\ &= 4 \end{aligned}

Now take the log base 4 of both sides. When you did that originally, I noticed that on the left side you had this:
log44 1/(x+1) · log48 1/(x+2)

You should use a single log for the product, instead of a log for each factor, like this:
$$\log_4 \left( 4^{1/(x + 1)} \cdot 8^{1/(x + 2)} \right)$$

Furthermore, $$\log_4 8$$ can be simplifed to a rational number:
$$\log_4 8 = \log_4 4^{3/2} = \frac{3}{2}$$

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Last edited: Aug 23, 2010
3. Aug 23, 2010

### ehild

Wrong. log(a*b)=log(a)+log(b).

ehild

4. Aug 23, 2010

### hunt_mat

Can't you write 4=2^2, and 8=2^3 to obtain:
$$2^{2/(1+x)}\cdot 2^{3/(x+2)}=3^{4\log 3}$$
Which when rearranged shows:
$$2^{2/(1+x)+3/(x+2)}=3^{4\log 3}$$
From here it looks straight forward to solve.

Mat

5. Aug 23, 2010

### ExamFever

Thank you all for the quick responses.

I have one more quick question about taking the logarithm of a side of an equation.

Consider the equation

32x+1 - 28 * 3x + 9 = 0

Now, it is obvious that a logarithm with base 3 needs to be taken, what I am wondering about is whether I have distributed the signs appropriately in the following calculation:

log332x+1 - (log328 + log33x) + log39 = 0

implies

2x + 1 - log328 - x + 2 = 0

implies the solution

x = log328 - 3

6. Aug 23, 2010

### hunt_mat

write u=3^{x}, then your equation reduces to:
$$3u^{2}-28u+9=0$$
Solve this and then take logs to base 3.

7. Aug 23, 2010

### ExamFever

I don't seem to follow you completely.

If 3x = u

How is 32x+1 = 3u2 ??

8. Aug 23, 2010

### hunt_mat

This use nothing but the laws of powers.
$$(3^{x})^{2}=3^{2\cdot x}=3^{2x}$$
Then:
$$3\cdot 3^{2x}=3^{2x+1}$$

9. Aug 23, 2010

### ExamFever

Ok got it...wow, I really should learn to take all rules into account, I find it hard to see those things straight away like you seem to be able to :S

Here is one last one:

lg(x2-5x+7) / lg(4-x) = 0

I assume with lg log10 is meant.

Because this is a fraction both the numerater and denominator may be multiplied by the same number, in this case:

log1010(x2-5x+7)/log1010(4-x)

Is this correct?

10. Aug 23, 2010

### hunt_mat

You have an equation:
$$\frac{\log(x^{2}-5x+7)}{\log (4-x)}=0$$
Multiply though by log (4-x), to obtain:
$$\log (x^{2}-5x+7)=0$$
So this means that:
$$x^{2}-5x+7=1\Rightarrow x^{2}-5x+6=0$$

11. Aug 23, 2010

### hunt_mat

I am teaching this sortof stuff at the moment, so don't feel too bad.

12. Aug 23, 2010

### ExamFever

Allright, one more and I should have a clear understanding of everything they can throw at me (hopefully).

32+log925 + 251-log52 + 10-lg4

Now, From the previous problem, I have a feeling that before doing anything to all of the terms, first the terms have to be simplified individually.

For the last one I can see that:

10-lg4 = 10lg4-1 = 4-1 = 1/4

For

32+log925

I can see there is a 32 in there somewhere, but how to get it out of the 2+log925?

Is

32(1+.5log925) = 91+.5log925

A valid way of doing this? If so, how to continue from here?

For

251-log52 = 52(1-log52)

And if I continue

52-2log52 = 52-log54 = 2-4 ???

What to do with those numbers that are not part of the logarithm?

13. Aug 23, 2010

### ExamFever

Is replacing 2 with the logarithm with base 5 that equals 2 a solution?

5log525-log54 = 5log5(25/4) = 25/4???

14. Aug 23, 2010

### hunt_mat

15. Aug 23, 2010

### ExamFever

Hmm and for

32+log925 = 32(1+.5log925) = 91+log95

This would mean:

9log99+log95 = 9log9(9*5) = 45

16. Aug 23, 2010

### eumyang

Note that
$$a^{m + n} = a^m \cdot a^n$$
and
$$b\log a = \log a^b$$

You'll need these rules to simplify further.

This is not correct. Use this rule:
$$a^{m - n} = \frac{a^m}{a^n}$$

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Last edited: Aug 23, 2010
17. Aug 23, 2010

### ExamFever

Ok I hope this is clearer:

10^-lg 4 = 10^lg 4-1 = 4-1 = 1/4

3^2+log925 = 3^2(1+.5 log925) = 9^1+log95 = 9^log99+log95 = 9^log9(9*5) = 45

18. Aug 23, 2010

### ExamFever

And for the last:

25^1-log52 = 5^2(1-log52) = 5^2-2 log52 = 5^2-log54 = 5^log525-log54 = 5^log5(25/4) = 25/4

19. Aug 23, 2010

### eumyang

Actually, not really. I'd learn LaTeX if I were you.
Or,
$$3^{2 + \log_9 25} = 3^{2(1 + 0.5\log_9 25)} = 9^{1 + 0.5\log_9 25} = 9^1 \cdot 9^{0.5\log_9 25} = 9 \cdot 9^{\log_9 5} = 9 \cdot 5 = 45$$

69

20. Aug 23, 2010

### ExamFever

Well, I'd rather spend time studying these problems at the moment, I have an exam to pass on friday...

Anyway, is it difficult? I mean if it's easy I might as well use it.

Anyhow, the last "translation" you did was what I wrote...is it correct?