How do you solve $\frac{log (35 - x^{3})}{log (5 - x)} = 3$ $(log(35 - x^3))/( log( 5 - x ))=3$ ?

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Answer 1 · 2016-07-17T17:13:18

The soln. is $x = 3, x = 2$ $x=3, x=2$ .

We will use these Rules :

$(R 1) : m log a = {log a}^{m}$ $(R1) : mloga=loga^m$

$(R 2) : {(a - b)}^{3} = a^{3} - b^{3} - 3 a b (a - b)$ $(R2) : (a-b)^3=a^3-b^3-3ab(a-b)$

Now, given that, $\frac{log (35 - x^{3})}{log (5 - x)} = 3$ $log(35-x^3)/log(5-x)=3$

$:. log(35-x^3)=3log(5-x)=log(5-x)^3$

Since log is $1-1$ fun., we get, $35-x^3=(5-x)^3$

$:. 35-x^3=125-x^3-3*5*x(5-x)$

$=125-x^3-75x+15x^2$

$:. 15x^2-75x+90=0$

$:. x^2-5x+6=0$

$:. (x-3)(x-2)=0$

The soln. is $x=3, x=2$ .

How do you solve (log(35 - x^3))/( log( 5 - x ))=3log(35−x3)log(5−x)=3(log(35 - x^3))/( log( 5 - x ))=3?