How do you solve $243^{.2 x} = 81^{x + 5}$ $243^(.2x)=81^(x+5)$ ?

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Answer 1 · 2016-03-11T13:45:48

$x = - \frac{20}{3}$ $x = -20 /3$

$243^{.2 x} = 81^{x + 5}$ $243 ^(.2x) = 81^ (x+5)$

Simplifying the bases by prime factorisation:

$243 = 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 3^{5}$ $243 =3*3*3*3*3 =color(blue)( 3^5$

$81 = 3 \cdot 3 \cdot 3 \cdot 3 = 3^{4}$ $81 = 3*3*3*3 = color(blue)(3^4$

Thus, using the rule ${(a^{b})}^{c} = a^{(b c)}$ $(a^b)^c=a^((bc))$ :

$243^{, 2 x} = 3^{(5 \times .2 x)} = 3^{x}$ $243 ^(,2x) = 3^((5xx.2x)) = 3^x$

$81^{x + 5} = 3^{(4 (x + 5))} = 3^{4 x + 20}$ $81^ (x+5) = 3^((4(x+5))) = 3^(4x + 20)$

The expression becomes:

$3^{x} = 3^{4 x + 20}$ $3^ (x) = 3^(4x + 20)$

The bases are equal so we can equate powers:

$x = 4 x + 20$ $x= 4x + 20$

$x - 4 x = 20$ $x-4x= 20$

$- 3 x = 20$ $-3x = 20$

$x = - \frac{20}{3}$ $x = -20/3$

How do you solve 243.2x=81x+5243^(.2x)=81^(x+5)?