Solve for ##m## in this equation that involves indices

[chwala]

Homework Statement

Find value of $m$, given $\frac {8^m + 27^m}{12^m + 18^m}$= $\frac {7}{6}$

Relevant Equations

indices and logs

Find value of $m$, given $\frac {8^m + 27^m}{12^m + 18^m}$= $\frac {7}{6}$

 

[chwala]

This is my working,
$\frac {2^{3m} + 3^{3m}}{(2 ⋅2 ⋅3)^m +(2⋅3⋅3)^m}$= $\frac {7}{6}$
$\frac {2^{3m} + 3^{3m}}{(2^{2m}⋅3^m + 2^m ⋅3^{2m}}$= $\frac {7}{6}$
now using the following properties,
$(a+b)(a-b)=a^2-b^2$
$(a^2-b^2)(a-b)$=$a^3-a^2b-ab^2+b^3$
$(a^2-b^2)(a-b)+a^2b+ab^2=a^3+b^3$

therefore, we shall have;
$\frac {(2^{2m} - 3^{2m})(2^m - 3^m)+2^{2m} 3^m +2^m3^{2m}}{(2^{2m}⋅ 3^m + 2^m⋅ 3^{2m}}$= $\frac {7}{6}$

$\frac {(2^{2m} - 3^{2m})(2^m - 3^m)+2^m 3^m(2^m +3^m)}{(2^m⋅ 3^m (2^m+ 3^m)}$= $\frac {7}{6}$

$\frac {(2^m + 3^m)(2^m-3^m)(2^m-3^m)+2^m3^m(2^m+3^m)}{(2^m ⋅3^m (2^m+ 3^m)}$= $\frac {7}{6}$

$\frac {(2^m - 3^m)(2^m-3^m)+2^m3^m}{2^m. 3^m }$= $\frac {7}{6}$
$6⋅2^{2m} + 6 ⋅3^{2m} - 6 ⋅2^m⋅3^m = 7⋅2^m⋅3^m$
$6⋅2^{2m} + 6⋅3^{2m} =13⋅2^m⋅3^m$
let $a=2^m$ and $b=3^m$ then it follows that,
$6a^2+6b^2=13ab$
$6a^2-13ab+6b^2=0$
$(2a-3b)(3a-2b)=0$
From,$2a-3b=0$,it follows that,
$\frac {a}{b}$=$1.5$
$\frac {2^m}{3^m}$=$1⋅5$ ,
$m$ =$-0.9975$

also from,
$3a-2b=0$,
$\frac {a}{b}$=$0.66666$
$\frac {2^m}{3^m}$=$0.66666$
$→m = 1$

 

[fresh_42]

It is always a good strategy to remove what disturbs. We only have factors $2^m$ and $3^m$. What do we get if we substitute $2^m=x$ and $3^m=y$? Can you simplify the equation in terms of $x$ and $y$?

 

[chwala]

It is always a good strategy to remove what disturbs. We only have factors $2^m$ and $3^m$. What do we get if we substitute $2^m=x$ and $3^m=y$? Can you simplify the equation in terms of $x$ and $y$?

allow me to finish posting my solution...

 

[chwala]

yap, looking for other ways of doing this...

 

[fresh_42]

I think you made a mistake after $(2a-3b)(3a-2b)=0.$

$(2a-3b)=0$ means $\dfrac{a}{b}=\dfrac{2^m}{3^m}=\dfrac{3}{2}$.
$(3a-2b)=0$ means $\dfrac{a}{b}=\dfrac{2^m}{3^m}=\dfrac{2}{3}$.

 

[chwala]

I think you made a mistake after $(2a-3b)(3a-2b)=0.$

$(2a-3b)=0$ is impossible. Why?
$(3a-2b)=0$ means $\dfrac{a}{b}=\dfrac{2}{3}$.

Why not? I think the solution which is approximately $m=-1$, satisfies the equation...we have two solutions here, $m=-1$ and $m=1$

 

[fresh_42]

My solution is basically the same:

$\dfrac {8^m + 27^m}{12^m + 18^m}= \dfrac {7}{6}\; , \;a=2^m\; , \;b=3^m$
\begin{align*}
\dfrac{a^3+b^3}{a^2b+ab^2}&=\dfrac{(a+b)(a^2-ab+b^2)}{ab(a+b)}=\dfrac{a^2-ab+b^2}{ab}=\dfrac{7}{6}\\
\dfrac{a}{b}-1+\dfrac{b}{a}&=c-1+\dfrac{1}{c}=\dfrac{7}{6}\text{ with }c=\dfrac{a}{b}\\
c^2-\dfrac{13}{6}c+1&=\left(c-\dfrac{2}{3}\right)\cdot \left(c - \dfrac{3}{2}\right)=0\\
\Longrightarrow \dfrac{a}{b}&=\left(\dfrac{2}{3}\right)^m \in \left\{\dfrac{2}{3}\, , \,\dfrac{3}{2}\right\}
\end{align*}

 

[fresh_42]

Why not? I think the solution which is approximately $m=-1$, satisfies the equation...

My mistake. I corrected it. $m=-1$ is indeed a solution.

 

[chwala]

My mistake. I corrected it. $m=-1$ is indeed a solution.

bingo Fresh