Logs!!..But no wood..- part 4

by observation since there is a root x in log therefore there can be no negative value for x and 0 cannot be a solution therefore the only options that remains is the first one :-)

For those wondering how to actually solve the equation manually...

Start by using the exponent rule for logs so that we get just $x$ inside each log:

$\frac{1}{2}*4\log_{\frac{x}{2}}(x) + 2*2\log_{4x}(x) = 3*3\log_{2x}(x)$

or,

$2\log_{\frac{x}{2}}(x) + 4\log_{4x}(x) = 9\log_{2x}(x)$

Since all of the coefficients of $x$ in the bases of each log are powers of 2, let's convert all of the logs to base 2 using the change of base formula:

$\frac{2\log_2(x)}{\log_2(\frac{x}{2})} + \frac{4\log_2(x)}{\log_2(4x)} = \frac{9\log_2(x)}{\log_2(2x)}$

Now we'll use the product and quotient rules for logs to simplify the denominators:

$\frac{2\log_2(x)}{\log_2(x)-\log_2(2)}$ + $\frac{4\log_2(x)}{\log_2(x)+\log_2(4)}$ = $\frac{9\log_2(x)}{\log_2(x)+\log_2(2)}$

which simplifies to:

$\frac{2\log_2(x)}{\log_2(x)-1}$ + $\frac{4\log_2(x)}{\log_2(x)+2}$ = $\frac{9\log_2(x)}{\log_2(x)+1}$

Let $y = \log_2(x)$ . Then by substitution, we have:

$\frac{2y}{y-1}$ + $\frac{4y}{y+2}$ = $\frac{9y}{y+1}$

Now let's get a common denominator on the left-hand side of the equation:

$\frac{2y(y+2)}{(y-1)(y+2)}$ + $\frac{4y(y-1)}{(y-1)(y+2)}$ = $\frac{9y}{y+1}$

$\frac{2y^2 + 4y + 4y^2 - 4y}{y^2+y-2} = \frac{9y}{y+1}$

$\frac{6y^2}{y^2+y-2}$ = $\frac{9y}{y+1}$

Cross-multiply and solve for $y$ :

$6y^2(y+1) = 9y(y^2+y-2)$

$6y^3 + 6y^2 = 9y^3 + 9y^2 - 18y$

$3y^3 + 3y^2 - 18y = 0$

$3y(y^2 + y - 6) = 0$

$3y(y+3)(y-2) = 0$

This tells us that $y$ can be 0, 2, or -3. Since $y = \log_2(x)$ , we have $x = 2^y$ , and then by substituting each value of $y$ , we find that $x$ can be either 1, 4, or $\frac{1}{8}$ .

A standard technique taught for people taking multiple-choice tests is to quickly survey the possible answered and see how far you can get by eliminating impossible solutions. Often this is easier to do than plowing through a difficult math problem. As Avn noted earlier, this problem asks to solve for x and gives us several complicated terms involving $\sqrt{x}$ and three logarithms with a bases of $x/2$ , $4x$ , and $2x$ respectively. The term involving the square root eliminates any solution suggesting a negative value of $x$ . And since there is no definition for the logarithm function with a base of 0, any solution involving 0 is also eliminated. And that leaves only one possibility: $1,4,1/8$ .

It is easy to see that $1$ is a solution, since the logarithm of $1$ is $0$ for any base.

For $4$ : we just verify that $4 \log_2 2 + 2 \log_{16} 16 = 3 \log_8 64$ ; i.e., $4+2 = 3*2$ .

For $1/8$ : $4 \log_{1/16} 2^{-3/2} + 2 \log_{1/2} (1/64) = 3 \log_{1/4} (1/512)$ , leads to $4 \log_{2^{-4}} 2^{-3/2} + 2 \log_{1/2} (1/64) = 3 \log_{2^{-2}} (2^{-9})$ , which simplifies to $(4 * (-3/2)) / (-4) + 2 * 6 = 3 * 9/2$ , or $1.5 + 12 = 13.5$ .

Here we use the general principle that $\log_{c^x} (c^y) = y/x$ for any non-zero values of $c$ , $x$ , and $y$ .