# Is there a way to solve these questions by using logarithm and without calculator?

$$2^{x+2} + 2^{x-1} + 2^x = 44$$

• Find the unknown $x$.

Indeed, I'm able to solve this question. However, there was something which they were doing the calculations by using logarithm. According to my views, it requires a calculator which we've to calculate with. Unfortunalety, we're not allowed to use calculator during the exam. Is there a way to solve these questions by using logarithm and without calculator?

My Kindest Regards!

• Write $y=2^x$. – Lord Shark the Unknown Jan 9 '18 at 20:07
• Hint: $2^{x+2} = 2(2^{x+1})$ – pureundergrad Jan 9 '18 at 20:08
• So, does log requires calculator? – user518016 Jan 9 '18 at 20:14
• @Enzo Taking logarithms does not require a calculator. You can either leave an answer simplified with a logarithm (for example $\log_2(3)$ would be a completely reasonable solution [though not to this problem]), or if the expression is straight-forward enough, you can simplify the log to something nicer (in this problem, the solution is $\log_2(8)$, which simplifies to $3$, as $8 = 2^3$). – Xander Henderson Jan 9 '18 at 20:19
• @Enzo Please, if you are ok, you can accept the answer and set it as solved. Thanks! – user Feb 4 '18 at 0:07

it is $$2^x\left(4+\frac{1}{2}+1\right)=44$$

If calculator is not required, then it needs to be easy to solve, with integer numbers. Try with $x=1,2,3$ $32 + 8 + 4$

$x=3$

$2^5 + 2^2 + 2^3$

• Have you solved it with log? – user518016 Jan 9 '18 at 20:13

As has been suggested by several people, the first simplification to make is to factor $2^x$ out of each term. This leave you with $$2^{x+2} + 2^{x-1} + 2^x = 2^x \left( 2^2 + 2^{-1} + 1 \right) = 2^x \left( 4 + \frac{1}{2} + 1 \right) = 2^x \cdot \frac{11}{2} = 44.$$ We can then cancel the $\frac{11}{2}$ in order to isolate the $2^x$ (which is the most complicated looking part of the equation at this point), which gives us $$2^x = 44\cdot \frac{2}{11} = 4\cdot 2 = 8.$$ Taking logarithms, we finally obtain $$x = \log_2(2^x) = \log_2(8) = \log_2(2^3) = 3.$$

• Can't we use log at begin? – user518016 Jan 9 '18 at 20:14
• @Enzo In what way? – Xander Henderson Jan 9 '18 at 20:15
• So we should use it at final, shouldn't we? – user518016 Jan 9 '18 at 20:16
• @Enzo There is no "ought to." If a technique works, great! I just don't see how taking logs without simplifying first is going to get us to a solution. But my lack of vision doesn't imply that there is no such technique. – Xander Henderson Jan 9 '18 at 20:16
• Which is more useful? taking logs or simpiflying? – user518016 Jan 9 '18 at 20:18

You can group by $2^x$ and rewrite it as: $$2^{x+2} + 2^{x-1} + 2^x = 2^x\left(2^2+2^{-1}+1\right)=44$$ $$2^x\cdot\frac{11}2=44$$ $$2^x=\frac{44\cdot2}{11}=8$$ $$2^x=2^3$$ Then you can get the logarithm of both LHS and RHS and you get: $$\log_22^x=\log_22^3\rightarrow\color{red}{x=3}$$

Were going to use the property of exponents $$2^{m+n}=2^m \cdot 2^n$$

The given equation, is this equivalent to $$2^x(4+\frac 12 +1)=44$$

$$\implies 2^x \times \frac{11}{2} =44 \implies 2^x =8$$

Since $8=2^3$ write $2^x=2^3$ and take $\log$ to the base $2$, giving $x=3$

Let $2^x=a$

thus

$$2^{x+2} + 2^{x-1} + 2^x = 44\iff4a+\frac{a}2+a=44\iff a\frac{11}{2}=44\iff a=8\iff x=3$$

By inspection (trial and error) $x=3$ is a solution. Since the LHS is strictly increasing it is unique.