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Solve. (6√6)^−x+3 = 1/6·216^2x−3

 

Step by step help please

 Jan 14, 2020
 #1
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Is it [(6√6)^-x]+3=1/6·[216^2x]−3

or (6√6)^[−x+3] = 1/6·216^[2x−3]

 
 Jan 14, 2020
 #2
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it's  (6√6)^[−x+3] = 1/6·216^[2x−3]

 
Guest Jan 14, 2020
 #3
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This problem may look it involves logs but It can be solved using exponentials and indices laws.

 

Let's first write it in latex form so it is clearer:

\((6*\sqrt{6})^{-x+3}=\frac{1}{6}*216^{2x-3}\)

Notice the bracket \((6*\sqrt{6})^{-x+3}\) Ignore the exponent right now, 

we know that: \((6*\sqrt{6})\)=\((6*6^{0.5})\) 

Now back the exponent

\((6*6^{0.5})^{-x+3}\) using the law: \((a*b)^c=a^c*b^c\)

\(6^{-x+3}*6^{0.5*(-x+3)}\) Multiply the bracket by the half, we get, \(6^{-x+3}*6^{-0.5x+1.5}\)

Using the law \(a^{b+c}=a^b*a^c\)

\(6^{-x}*6^3*6^{-0.5x}*6^{1.5}\) 

The right hand side now,

\(\frac{1}{6}*216^{2x-3}\) 

We know that 

\(216=6^3\)

\(\frac{1}{6}=6^{-1}\)

\(6^{-1}*6^{3(2x-3)}\) apply the same rules,

\(6^{-1}*6^{6x-9}\) 

Now we have both sides simplified 

\(6^{-x}*6^3*6^{-0.5x}*6^{1.5}\)\(=\)\(6^{-1}*6^{6x-9}\)

Notice -x and -0.5x same base, we add exponents 

Notice 3 and 1.5, same base again so we add exponents

\(6^{-1.5x}*6^{4.5}\)\(=\)\(6^{-1}*6^{6x-9}\)

 

Now divide by \(6^{-1}\)

\(6^{-1.5x}*6^{4.5-(-1)}\)\(=\)\(6^{-1.5x}*6^{5.5}=6^{6x-9}\)

Notice \(6^{6x-9}=6^{6x}*6^{-9}\)

\(6^{-1.5x}*6^{5.5}=6^{6x}*6^{-9}\)

Divide by \(6^{-9}\)

\(6^{-1.5x}*6^{14.5}=6^{6x}\)

Divide by \(6^{-1.5x}\)

\(6^{14.5}=6^{7.5x}\)

The same base, therefore, exponents are equal (Law: \(a^b=a^c,b=c\))

\(14.5=7.5x\) Divide by 7.5

\(x=\frac{14.5}{7.5}=1.93333333\) the 3 is recurring.

 
 Jan 14, 2020

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