1. ## Logarithm Problems

Can someone help me?

1. $\log_{4}x - \log_{16}(x+3) = \frac{1}{2}$
2. $\log(x+12) + \log(x-3) = 2$
3. $\log_{x^2}9 - \log_{x}36 = 1$

And for this problem, can you check if I solved it right:
$\log_{x}2 + \log_{x}3 = 5$
$\log_{x}(2 * 3) = 5$
$\log_{x}6 = 5$
$6 = x^5$
$1.431 = x$

Thank you

2. Originally Posted by meiyukichan
Can someone help me?
[*] $\log_{4}x - \log_{16}(x+3) = \frac{1}{2}$
$\log_{16}x= y$ mean $x= 16^y= (4^2)^y= 4^{2y}$ or $2y= \log_{4}x$ and $y=\log_{16}(x)=\log_4x/2$

You can write this as $\log_{4}x-\frac{1}{2}\log_4(x+3)= \log_4 x(x+3)^{1/2}= \frac{1}{2}$

$x(x+3)^{1/2}= 4^{1/2}= 2$.

Squaring both sides, $x^2(x+ 3)= 4$, $x^3+ 3x^2- 4= 0$. By the "rational root theorem" the only possible rational roots are x= 1, x= -1, x= 2, x= -2, x= 4, and x= -4. Try those to see if any are actually roots.

[*] $\log(x+12) + \log(x-3) = 2$
Almost trivial: $\log((x+ 12)(x-3))= 3$. Assuming that "log" here means common logarithm, $(x+12)(x-3)= 10^2= 100$. Solve that quadratic equation.

[*] $\log_{x^2}9 - \log_{x}36 = 1$
Similar to the first one isn't it? $\log_{x^2}9= y$ means $9= (x^2)^y= x^{2y}$ so $2y= log_x9$ and $y= \frac{1}{2}log_x9= log_x 9^{1/2}= log_x 3$.

$\log_x 3- \log_x 36= \log_x \frac{3}{36}= 1$ so $\frac{3}{36}= \frac{1}{12}= x^1$.

And for this problem, can you check if I solved it right:
$\log_{x}2 + \log_{x}3 = 5$
$\log_{x}(2 * 3) = 5$
$\log_{x}6 = 5$
$6 = x^5$
$1.431 = x$
Yes, that is correct.

[/quote]Thank you [/QUOTE]

3. Originally Posted by meiyukichan
$\log_{4}x - \log_{16}(x+3) = \frac{1}{2}$
Note that 16 = 4^2. Apply the change-of-base formula to log_16(x) to convert and get:

. . . . . $2\log_{16}(x)\, -\, \log_{16}(x\, +\, 3) =\, \frac{1}{2}$

. . . . . $\log_{16}(x^2)\, -\, \log_{16}(x\, +\, 3)\, =\frac{1}{2}$

. . . . . $\log_{16}\left(\frac{x^2}{x\, +\, 3}\right)\, =\, \frac{1}{2}$

Note that log_16(4) = 1/2, so:

. . . . . $\log_{16}\left(\frac{x^2}{x\, +\, 3}\right)\, =\, \log_{16}(4)$

Then:

. . . . . $\frac{x^2}{x\, +\, 3}\, =\, 4$

Solve the rational equation.

Originally Posted by meiyukichan
$\log(x+12) + \log(x-3) = 2$
A straight-forward application of log rules and the definition of the common log leads to:

. . . . . $\log\left((x\, +\, 12)(x\, -\, 3)\right)\, =\, \log(100)$

Equate the arguments, and solve the resulting quadratic equation.

Originally Posted by meiyukichan
$\log_{x^2}9 - \log_{x}36 = 1$
Note that the change-of-base formula says that log_{x^2}(9) = log_x(9) / log_x(x^2}, and log_x(x^2) obviously equals 2. Then:

. . . . . $\frac{\log_x(9)}{2}\, -\, \log_x(36)\, =\, \log_x(x)$

. . . . . $\log_x(\sqrt{9})\, -\, log_x(36)\, =\, \log_x(x)$

. . . . . $\log_x\left(\frac{3}{36}\right)\, =\, \log_x(x)$

Equate the arguments, and simplify to solve.

Originally Posted by meiyukichan
And for this problem, can you check if I solved it right:

$\log_{x}2 + \log_{x}3 = 5$
$\log_{x}(2 * 3) = 5$
$\log_{x}6 = 5$
$6 = x^5$
$1.431 = x$
Looks good to me! :wink: