1. ## log/exponent problem

The question consists of one line only, but the silly me failed to solve it.
Please find the value of x for this equation:

2. I cannot say for anyone else, but I have absolutely no idea what that says.
For example, what could $^4\log 2$ mean?
Let alone all the rest of the jumble of notation.

Why not learn to post in symbols? You can use LaTeX tags
$$\log_{4}(2)$$ gives $\log_{4}(2)$

3. aww, I am terribly sorry.
In my stupid country, $\log_{n}$ is written as $^{n}log$ instead.

So, it should be written like this for international standard:

$\frac{6}{5}(h^{\log_{h}x.\log_{10}h.\log_{h}5})-3^{\log\frac{x}{10}}=9^{\log_{100}x+\log_{4}2}$

4. Originally Posted by eternia
aww, I am terribly sorry.
In my stupid country, $\log_{n}$ is written as $^{n}log$ instead.So, it should be written like this for international standard:
$\frac{6}{5}(h^{\log_{h}x.\log_{10}h.\log_{h}5})-3^{\log\frac{x}{10}}=9^{\log_{100}x+\log_{4}2}$
I am sorry to tell you, that reply is useless.
What does tho O.P. mean?
What was included in to O.P. is gibberish.
What do you mean?

5. OK so I got the problem as:

Find the value(s) of x:

$\displaystyle \frac{6(h^{\log_{h}(x) \, \cdot \, \log_{10}(h) \, \cdot \, \log_{h}(5)})}{5} - 3^{\log(\frac{x}{10})}=9^{\log_{100}(x)+\log_{4}(2 )}$

For the $3^{\log\frac{x}{10}}$ part, what is the base of this logarithm? Is it supposed to be $log_e$?

To solve for x, notice how the h's can cancel out. The first step would be to cancel out the h's so you don't have 2 variables.

6. The one above me figured out my equation properly,
but for that one, the base number is 10, not e
If it's e, it would be written as Ln instead, no?
So, it's :

$3^{\log_{10}(\frac {x} {10})$

For starting, we cancel out the h of course, but I am still having difficulties....
You see, even though the 9 can be turned into $3^2$
But I can't find a way to trash the 5, which is a prime number on it's own.

7. I don't think you can solve this algebraically.

Maybe you could use an approximation method to solve for x instead?

8. My prof. said that it can be solved, x = 100
Our task is to show him the process....

9. This is quite tricky.
Split this equation up into 3 pieces:

$\displaystyle \underbrace{\tfrac{6}{5}(h^{\log_{h}(x) \, \cdot \, \log_{10}(h) \, \cdot \, \log_{h}(5)})}_{\text{Part a}} - \underbrace{3^{\log_{10}(\frac{x}{10})}}_\text{Par t b}}=\underbrace{9^{\log_{100}(x)+\log_{4}(2)}}_{\t ext{Part c}}$

I'll skip out some parts or else it will be too long.
Part a can be expressed as:

$\displaystyle \tfrac{6}{5}(h^{\log_{h}(x) \, \cdot \, \log_{10}(h) \, \cdot \, \log_{h}(5)})} = 6 \times 5^{\log_{10}(x) - 1}$

Part b can be expressed as:

$3^{\log_{10}(\frac{x}{10})} = 3^{\log_{10}(x) - 1}$

Part c can be expressed as:

$9^{\log_{100}(x)+\log_{4}(2)} = 3^{\log_{10}(x) + 1}$

Hence the equation becomes:

$\displaystyle 6 \times 5^{\log_{10}(x) - 1} -3^{\log_{10}(x) - 1} = 3^{\log_{10}(x) + 1}$

$\displaystyle 6 \times 5^{\log_{10}(x) - 1} = \underbrace{3^{\log_{10}(x) + 1} + 3^{\log_{10}(x) - 1}}_{\text{Simplify}}$

To simplify:

$3^{\log_{10}(x)} \times 3^1 + 3^{\log_{10}(x)} \times 3^{-1}$

$3^{\log_{10}(x)} ( 3^1 + 3^{-1})$

$3^{\log_{10}(x)} (\frac{10}{3}) \underbrace{\times 3 \div 3}_{\text{Multiply by 1}}$

$3^{\log_{10}(x)} \div 3 \times (\frac{10}{3}) \times 3$

$3^{\log_{10}(x)-1} \times 10$

Thus the equation becomes:

$\displaystyle 6 \times 5^{\log_{10}(x) - 1} =10 \times 3^{\log_{10}(x)-1}$

To make it easier, let $u = \log_{10}(x)$

$\displaystyle 6 \times 5^{u - 1} =10 \times 3^{u-1}$

Now use logarithms to solve for u, the thus solve for x.

10. Eureka!
I am forever indebted to you, big bro.
Thank you, thank you so much.