Q1 - Equation with fractional and logarithmic exponents

[ExamFever]

Homework Statement

4 ^1/(x+1) · 8 ^1/(x+2) = 9^{log3 2}

Homework Equations

n/a - same as 1

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!

Thanks in advance,

 

[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}<br /> 9^{\log_3 2} &amp;= (3^2)^{\log_3 2} \\<br /> &amp;= 3^{2 \log_3 2} \\<br /> &amp;= 3^{\log_3 4} \\<br /> &amp;= 4<br /> \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}69

 

[ehild]

Homework Statement

4 ^1/(x+1) · 8 ^1/(x+2) = 9^{log3 2}
...
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}

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

ehild

 

[hunt_mat]

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

Mat

 

[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

3^2x+1 - 28 * 3^x + 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:

log₃3^2x+1 - (log₃28 + log₃3^x) + log₃9 = 0

implies

2x + 1 - log₃28 - x + 2 = 0

implies the solution

x = log₃28 - 3

 

[hunt_mat]

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

 

[ExamFever]

I don't seem to follow you completely.

If 3^x = u

How is 3^2x+1 = 3u² ??

 

[hunt_mat]

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

 

[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?

 

[hunt_mat]

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

 

[hunt_mat]

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

 

[ExamFever]

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

3^2+log₉25 + 25^1-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 = 10^lg4-1 = 4^-1 = 1/4

For

3^2+log₉25

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

Is

3^{2(1+.5log₉25)} = 9^1+.5log₉25

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

For

25^1-log₅2 = 5^2(1-log₅2)

And if I continue

5^2-2log₅2 = 5^2-log₅4 = 2-4 ?

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

 

[ExamFever]

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

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

 

[hunt_mat]

Sort of, I can't quite read your equations, can you make then a little clearer please.

 

[ExamFever]

Hmm and for

3^2+log₉25 = 3^{2(1+.5log₉25)} = 9^1+log₉5

This would mean:

9^{log₉9+log₉5} = 9^log9(9*5) = 45

Please tell me I'm right...

 

[eumyang]

For

3^2+log₉25

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

Is

3^{2(1+.5log₉25)} = 9^1+.5log₉25

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

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.

For

25^1-log₅2 = 5^2(1-log₅2)

And if I continue

5^2-2log₅2 = 5^2-log₅4 = 2-4 ?

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

This is not correct. Use this rule:
a^{m - n} = \frac{a^m}{a^n}69

 

[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

 

[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

 

[eumyang]

Actually, not really. I'd learn LaTeX if I were you.

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

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

 

[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?

 

[eumyang]

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

I cannot speak for everyone here, but my point was that some people are less likely to help you if they cannot read your problem/work.

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

What I posted previously was not a "translation" of what you wrote, but an alternate way of arriving at the solution. You rewrote 1 as \log_9 9, while I used the product of powers property a^{m + n} = a^m \cdot a^n. You get the same answer either way.69

 

[ExamFever]

Okay, I understand that.

The "translation" part was more of a joke, you translated my bad writing into latex ;)

Thank you very much for the help, it is greatly appreciated!

 

[vela]

It's not so important for you to learn LaTeX for this type of expression. You can get superscripts and subscripts using, respectively, the sup and sub tags. It's more important that you write it correctly. For example, you wrote 25^1-log₅ 2. According to the standard rules of precedence, this is actually equal to 25-log₅ 2 = 25-0.431 = 24.569, not 25/4. What you meant was

25^(1-log₅ 2)

or

25^{1-log₅ 2}.

 

[ExamFever]

I've made very good progress in solving logarithmic equations. I thought I had all of them covered until I ran into the following problem:

log₂(x+1) + log_x+12 = 5/2

I've tried several things here, but never managed to work out all logarithms.

Some equations up to where I got and feel I'm in a right place:

x+1=u

u² + log_u16 = 32

Worked away the first logarithm, got an integer number on the RHS, but still this annoying logarithm!

u² + 4 = 2^u5/2

All logarithms are gone, but in order to get that u^5/2 down we'll have to introduce logarithms again...

I'm running in circles here. Help would be greatly appreciated!

 

[hunt_mat]

<br /> \log_{a}b=\frac{1}{\log_{b}a}<br />

 

[ExamFever]

That is very good to know. So when working away logarithms, how will the fraction be treated?

log₂(x+1) + (1 / log₂(x+1)) = 5/2
x + 1 + (1 / (x+1)) = 4 sqrt(2

Is this correct?

 

[hunt_mat]

<br /> u=\log_{2}(x+1)<br />
Then:
<br /> u+\frac{1}{u}=\frac{5}{2}<br />