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My second question for the day is this:

\(2^{x+3}-3*2^{x-1}=104\)

I don't understand how one would go about the 3 which is multiplied with an exponential number. I'm going to be bold and place the last sum I have here as well, because it is very similar to the one on top, If you want, please show me the top one only, and I'll learn from that and try myself to do the second one, otherwise if you see a "catch" maybe in the second one, please show me?..

\(2^{2x}+7*2^x=8\)

thank you all very very much!!

juriemagic Feb 6, 2019

#1**+1 **

\(2^{x+3}-3\cdot 2^{x-1} = 104\\ 8\cdot 2^x - \dfrac 3 2 2^x = 104\\ \dfrac{13}{2}\cdot 2^x = 104 \\ 2^x = 16\\ x = 4\)

The second one is a tad trickier

\(2^{2x} + 7 \cdot 2^x = 8\\ u=2^x\\ u^2 + 7u - 8 = 0\\ (u+8)(u-1) = 0\\ u = -8, 1\\ x = \log_2(u)\\ \text{-8 is seen to be an extraneous solution}\\ x = \log_2(1)=0\\ \text{and indeed}\\ 2^0 + 7 \cdot 2^0 = 1+7=8\)

.Rom Feb 6, 2019

#2**+1 **

goodness me!!,

Rom, every year the math is becoming more and more complicated...this is for a grade 11 pupil...I cannot believe the complexity of the sums nowadays...thank you very much Rom, I definitely would never have been able!..thank you!!

juriemagic
Feb 6, 2019

#3**+9 **

**just algebra q2**

**\(2^{2x}+7*2^x=8\)**

\(\begin{array}{|lrcll|} \hline & 2^{2x}+7*2^x &=& 8 \\ & 2^{2x}+7*2^x -8 &=& 0 \\ & (2^x+8)(2^x-1) &=& 0 \\\\ 1. & 2^x+8 &=& 0 \\ & 2^x &=& -8 \quad x \text{ is an imaginary number},\ x = \dfrac{2 i \pi n + i \pi + 3 \ln(2)}{\ln(2)}, n \in \mathbb{ Z} \\\\ 2. & 2^x-1 &=& 0 \\ & 2^x &=& 1 \\ & 2^x &=& 2^0 \\ & \mathbf{x} & \mathbf{=} & \mathbf{0} \\ \hline \end{array} \)

heureka Feb 6, 2019