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$$\frac {(2^{a+3}+ 2^{a+1}) \cdot 2^{a+2}}{2^{a+3}} = 20$$

  • Find the unknown $a$.

So, we began learning exponential equations. However, I want to solve this problem by simpifyling it.

Regards

  • Use the substitution $x=2^a$. – Mengchun Zhang Jan 09 '18 at 17:38
  • @MengchunZhang How? –  Jan 09 '18 at 17:38
  • For example, $2^{a+3}=2^3\cdot 2^a=8x$. – Mengchun Zhang Jan 09 '18 at 17:39
  • @MengchunZhang Where that $8x$ come from? –  Jan 09 '18 at 17:40
  • Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Fly by Night Jan 09 '18 at 17:41
  • @FlybyNight Okay, If you want me to show my effort: $$(2^a . 2^3 . 2^a . 2) . 2^a . 2^2$$ Hence, $$2^a(2^3 . 2)$$ and $$2^a . 16$$. –  Jan 09 '18 at 17:41
  • @Enzo That should be included in the body of your question. People will be much more willing to help if you show your working. Just some friendly advice :-) (People on the site get annoyed when people just ask for answers, like we're some kind of free homework service.) – Fly by Night Jan 09 '18 at 18:09

3 Answers3

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What did you already try? Because I think it's good to start to work out the parentheses ( ), for example: $(a + b) \cdot c = ac + bc$

MathMe
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HINTS

Recall the power rules: $2^p \times 2^q = 2^{p+q}$ and $2^p \div 2^q = 2^{p-q}$.

Expand the numerator.

Divide each term in the (expanded) numerator by the denominator.

Take out a common factor on the left, and divide both sides by that common factor.

Solve for $a$.

Fly by Night
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$$\left(2^{\left(a+3\right)}+2^{\left(a+1\right)}\right)\cdot\frac{2^{\left(a+2\right)}}{2^{\left(a+3\right)}}$$ $$\left(2\cdot2\cdot2^{\left(a+1\right)}+2^{\left(a+1\right)}\right)\cdot\frac{2^{\left(a+1\right)}\cdot2}{2^{\left(a+1\right)}\cdot2\cdot2}$$ $$\left(2\cdot2\cdot2^{\left(a+1\right)}+2^{\left(a+1\right)}\right)\cdot\frac{2}{2\cdot2}$$ $$5\cdot2^{\left(a+1\right)}\cdot\frac{1}{2}$$

Now let's set this equal to 20: $$5\cdot2^{\left(a+1\right)}\cdot\frac{1}{2}=20$$ $$2^{\left(a+1\right)}=8$$ $$a=2=\log_2\left(8\right)-1$$

Finn Eggers
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