# Math Help - Logrithms

1. ## Logrithms

This is a problem i'm having trooble on
i put it as an attachment

thanks

2. Hello, anime_mania!

$\text{(a) If }\left(\log_3x\right)\left(\log_x2x\right)\left(\l og_{2x}y\right) \;=\;\log_xx^2\text{, find the value of }y.$
On the right side: . $\log_xx^2 \:=\:2\!\cdot\log_xx \:=\:2\cdot1 \:=\:2$

Use the Base-Change Formula: . $\log_b(x) \:=\:\frac{\log(x)}{\log(b)}$

Then: . $\log_3x \:=\: \frac{\log x}{\log 3} \qquad \log_x2x \:=\: \frac{\log2x}{\log x} \qquad \log_{2x}y \:=\: \frac{\log y}{\log2x}$

The equation becomes: . $\frac{\log x}{\log 3}\cdot\frac{\log2x}{\log x}\cdot\frac{\log y}{\log2x} \:=\:2$

. . which simplifies to: . $\frac{\log y}{\log 3} \:=\:2\quad\Rightarrow\quad \log y \:=\:2\!\cdot\!\log3 \:=\:\log3^2$

Therefore: . $\log y \:=\:\log9\quad\Rightarrow\quad\boxed{y \:=\:9}$

$\text{(b) Find integral values of }x,y,z\text{ that satisfy all of the following equations.}$

. . ${\color{blue}[1]}\;\;z^x \;=\;y^{2x}$
. . ${\color{blue}[2]}\;\;2^z \;=\;2\cdot4^x$
. . ${\color{blue}[3]}\;\;x + y + z \;=\;16$

From [2], we have: . $2^z \:=\:2\cdot(2^2)^x\:=\:2\cdot2^{2x}\:=\:2^{2x+1}$
. . Hence: . $2x + 1 \:=\:z\quad\Rightarrow\quad x \:=\:\frac{z-1}{2}$ .[4]

From [1], we have: . $z^x \:=\:(y^2)^x\quad\Rightarrow\quad z = y^2$ .[5]

Substitute [5] into [4]: . $x \:=\:\frac{y^2-1}{2}$ .[6]

Substitute [5] and [6] into [3]: . $\frac{y^2-1}{2} + y + y^2 \:=\:16$

. . which simplifies to the quadratic: . $3y^2 + 2y - 33 \:=\:0$

. . which factors: . $(y - 3)(3y + 11) \:=\:0$

. . and has the integral solution: . $\boxed{y \:=\:3}$

Substitute into [6]: . $x \:=\:\frac{3^2-1}{2}\quad\Rightarrow\quad\boxed{x \:=\:4}$

Substitute into [5]: . $z \:=\:3^2\quad\Rightarrow\quad\boxed{z \:=\:9}$