Solve algebraically, showing each step of your working, the equation ${(8^{x - 1})}^{2} - 18 (8^{x - 1}) + 32 = 0$ $(8^(x-1) )^2-18(8^(x-1) )+32=0$ ??

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Answer 1 · 2018-02-24T21:05:36

$x = \frac{4}{3}$ $x=4/3$ or $x = \frac{7}{3}$ $x=7/3$

What we have here is a quadratic equation in terms of $8^{x - 1}$ $8^(x-1)$

Let $u = 8^{x - 1}$ $u=8^(x-1)$

The equation becomes

$u^{2} - 18 u + 32 = 0$ $u^2-18u+32=0$

$(u - 16) (u - 2) = 0$ $(u-16)(u-2)=0$

$u = 16$ $u=16$ or $u = 2$ $u=2$

$8^{x - 1} = 2$ $8^(x-1)=2$ or $8^{x - 1} = 16$ $8^(x-1)=16$

Considering $8^{x - 1} = 2$ $8^(x-1)=2$ first;

In log form;

${log}_{8} (2) = x - 1$ $log_8(2)=x-1$
$x - 1 = \frac{1}{3}$ $x-1=1/3$
$x = \frac{4}{3}$ $x=4/3$

Consider now $8^{x - 1} = 16$ $8^(x-1)=16$

In log form;

${log}_{8} (16) = x - 1$ $log_8(16)=x-1$
$x - 1 = \frac{4}{3}$ $x-1=4/3$
$x = \frac{7}{3}$ $x=7/3$

So $x = \frac{4}{3}$ $x=4/3$ or $x = \frac{7}{3}$ $x=7/3$

Solve algebraically, showing each step of your working, the equation (8^(x-1) )^2-18(8^(x-1) )+32=0(8x−1)2−18(8x−1)+32=0(8^(x-1) )^2-18(8^(x-1) )+32=0??