Question

.
${8}^{x + 2 } = {(16)}^{2x - 9}$
​

Answer 1

Answer:

$8^{x+2}=\left(16\right)^{2x-9}\quad :\quad x=\dfrac{42}{5}\quad \left(\mathrm{Decimal}:\quad x=8.4\right)$

Step-by-step explanation:

$8^{x+2}=\left(16\right)^{2x-9}$

$\mathrm{Convert\:}8^{x+2}\mathrm{\:to\:base\:}2$

$8^{x+2}=\left(2^3\right)^{x+2}$

$\left(2^3\right)^{x+2}=16^{2x-9}$

$\mathrm{Convert\:}16^{2x-9}\mathrm{\:to\:base\:}2$

$16^{2x-9}=\left(2^4\right)^{2x-9}$

$\left(2^3\right)^{x+2}=\left(2^4\right)^{2x-9}$

$\mathrm{Apply\:exponent\:rule}:\quad \left(a^b\right)^c=a^{bc}$

$\left(2^3\right)^{x+2}=2^{3\left(x+2\right)},\:\space\left(2^4\right)^{2x-9}=2^{4\left(2x-9\right)}$

$2^{3\left(x+2\right)}=2^{4\left(2x-9\right)}$

$\mathrm{If\:}a^{f\left(x\right)}=a^{g\left(x\right)}\mathrm{,\:then\:}f\left(x\right)=g\left(x\right)$

$3\left(x+2\right)=4\left(2x-9\right)$

$\mathrm{Solve\:}\:3\left(x+2\right)=4\left(2x-9\right)$

$\mathrm{Expand\:}3\left(x+2\right)$

$\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac$

$a=3,\:b=x,\:c=2$

$=3x+3\cdot \:2$

$\mathrm{Multiply\:the\:numbers:}\:3\cdot \:2=6$

$=3x+6$

$\mathrm{Expand\:}4\left(2x-9\right)$

$\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac$

$a=4,\:b=2x,\:c=9$

$=4\cdot \:2x-4\cdot \:9$

$\mathrm{Simplify}\:4\cdot \:2x-4\cdot \:9$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:2=8$

$=8x-4\cdot \:9$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:9=36$

$=8x-36$

$3x+6=8x-36$

$\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides}$

$3x+6-6=8x-36-6$

$\mathrm{Simplify}$

$3x=8x-42$

$\mathrm{Subtract\:}8x\mathrm{\:from\:both\:sides}$

$3x-8x=8x-42-8x$

$\mathrm{Simplify}$

$-5x=-42$

$\mathrm{Divide\:both\:sides\:by\:}-5$

$\dfrac{-5x}{-5}=\dfrac{-42}{-5}$

$\mathrm{Simplify}$

$x=\dfrac{42}{5}$