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avatar+2547 

I'm not that good at math, as I am only in middle school, so please help.

 

Express the following in simplest form : \((-2)^{4x+2}(2)^{6x-5}(8)^x\)

 

Please don't give me the answer, give me any hints about helpful strategies I can use to solve this.

 

Source: Mathcounts Lectures (18) Simplifying Algebra Expressions

 

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I simplified it to \(2^{18x^2-15x}*(-2)^{4x+2}\)

 

But I'm not sure how the -2 will affect the final simplified form

 

I have vague ideas that \((-2)^{4x+2}\) will always be positive,

 

My answer of \(2^{(18x^2-15x)(4x+2)}\) (haven't simplified binomials) doesn't seem to be quite correct...

 Apr 25, 2019
edited by CalculatorUser  Apr 25, 2019
edited by CalculatorUser  Apr 25, 2019
edited by CalculatorUser  Apr 25, 2019
 #1
avatar+19782 
+1

hint    :      3^x   *    3^4   =  3^(x+4)   not   3^(4x)

 Apr 25, 2019
 #2
avatar+23903 
+3

Express the following in simplest form : \((-2)^{4x+2}(2)^{6x-5}(8)^x\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{(-2)^{4x+2}(2)^{6x-5}(8)^x} \\ &=& \Big((-1)(2) \Big)^{4x+2}(2)^{6x-5}(8)^x \quad & | \quad \boxed{(ab)^c = a^cb^c} \\ &=& \left(-1 \right)^{4x+2}\left(2 \right)^{4x+2}(2)^{6x-5}(8)^x \quad & | \quad \boxed{a^{b+c} = a^ba^c} \\ &=& \left(-1 \right)^{4x}\left(-1 \right)^{2}\left(2 \right)^{4x+2}(2)^{6x-5}(8)^x \quad & | \quad \boxed{ (-1)^2 = (-1)(-1)=1 } \\ &=& \left(-1 \right)^{4x}\left(2 \right)^{4x+2}(2)^{6x-5}(8)^x \quad & | \quad 8 = 2^3 \\ &=& \left(-1 \right)^{4x}\left( 2 \right)^{4x+2}(2)^{6x-5} \left(2^3 \right)^x \quad & | \quad \boxed{ \left(a^b \right)^c = a^{bc} \\ \left(2^3 \right)^x = 2^{3x} } \\ &=& \left(-1 \right)^{4x}\left( 2 \right)^{4x+2}(2)^{6x-5} (2^{3x}) \quad & | \quad \boxed{ a^ba^ca^d=a^{b+c+d} } \\ &=& \left(-1 \right)^{4x}2^{4x+2+6x-5+3x} \\ &=& \left(-1 \right)^{4x}2^{4x+2+6x-5+3x} \\ &\mathbf{=}& \mathbf{\left(-1 \right)^{4x}2^{13x-3}} \\ \hline \end{array} \)

 

laugh

 Apr 25, 2019
edited by heureka  Apr 25, 2019
 #3
avatar+2547 
+3

wow, I totally messed up the exponent rules!!!

 

Thanks, I thought this would be a hard problem.

 Apr 25, 2019

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