How do you solve $3^{x + 1} \cdot 2^{x - 2} = 21$ $3^(x+1) * 2^(x-2)= 21$ ?

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Answer 1 · 2015-11-15T02:14:39

Use law of exponents to simplify

$3^{x} 3 \times \frac{2^{x}}{2^{2}} = 21$ $3^x3xx2^x/2^2=21$

${(3 \times 2)}^{x} (\frac{3}{4}) = 21$ $(3xx2)^x(3/4)=21$

$6^{x} = 28$ $6^x=28$

Next, use the property of logs ...

$x log 6 = log 28$ $xlog6=log28$

$x = \frac{log 28}{log 6} \approx 1.86$ $x=log28/log6~~1.86$

hope that helped

How do you solve 3^(x+1) * 2^(x-2)= 213x+1⋅2x−2=213^(x+1) * 2^(x-2)= 21?