Solve for X if $3^{x + 5} = 27^{- 2 x + 1}$ $3^(x+5)=27^(-2x+1)$ ?

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Answer 1 · 2018-05-30T08:36:44

$x = - \frac{2}{7}$ $x=-2/7$

$3^{x + 5} = 27^{- 2 x + 1}$ $color(blue)(3^(x+5)=27^(-2x+1)$

To solve questions like this, we need to make the bases same.

$\to 3^{x + 5} = {(3 \cdot 3 \cdot 3)}^{- 2 x + 1}$ $rarr3^(x+5)=(3*3*3)^(-2x+1)$

$\to 3^{x + 5} = {(3^{3})}^{- 2 x + 1}$ $rarr3^(x+5)=(3^3)^(-2x+1)$

Use, ${(a^{b})}^{c} = (a^{b c})$ $color(brown)((a^b)^c=(a^(bc))$

$\to 3^{x + 5} = 3^{- 6 x + 3}$ $rarr3^(x+5)=3^(-6x+3)$

Since, the bases are same, if $a^{b} = a^{c}$ $color(brown)(a^b=a^c$ then $b = c$ $color(brown)(b=c$

So,

$\to x + 5 = - 6 x + 3$ $rarrx+5=-6x+3$

$\to x + 6 x = 3 - 5$ $rarrx+6x=3-5$

$\to 7 x = - 2$ $rarr7x=-2$

$\Rightarrow x = - \frac{2}{7}$ $color(green)(rArrx=-2/7$

Hope that helps!... $ϕ$ $phi$

Solve for X if 3^(x+5)=27^(-2x+1)3x+5=27−2x+13^(x+5)=27^(-2x+1)?