Question
Solve for $x:2^{x+3}+2^{x+1}=320$
✓
Answer
$2^{x+3}+2^{x+1}=320$
$\Rightarrow 2^{x+3}+2^{x+1}=2^6 \times 5$
$\Rightarrow 2^x \cdot 2^3+2^x \cdot 2^1=2^6 \times 5$
$\Rightarrow 2^x\left(2^3+2^1\right)=2^6 \times 5$
$\Rightarrow 2^x(8+2)=2^6 \times 5$
$\Rightarrow 2^x(10)=2^6 \times 5$
$\Rightarrow 2^x\left(\frac{10}{5}\right)=2^6$
$\Rightarrow 2^x \cdot 2^6$
$\Rightarrow \frac{2^x \cdot 2}{2^6}=1$
$\Rightarrow 2^{x+1-6}=1 \times 2^0$
$\Rightarrow 2^{x-5}=1 \times 2^0$
$\Rightarrow x-5=0$
$\Rightarrow x=5$
$\Rightarrow 2^{x+3}+2^{x+1}=2^6 \times 5$
$\Rightarrow 2^x \cdot 2^3+2^x \cdot 2^1=2^6 \times 5$
$\Rightarrow 2^x\left(2^3+2^1\right)=2^6 \times 5$
$\Rightarrow 2^x(8+2)=2^6 \times 5$
$\Rightarrow 2^x(10)=2^6 \times 5$
$\Rightarrow 2^x\left(\frac{10}{5}\right)=2^6$
$\Rightarrow 2^x \cdot 2^6$
$\Rightarrow \frac{2^x \cdot 2}{2^6}=1$
$\Rightarrow 2^{x+1-6}=1 \times 2^0$
$\Rightarrow 2^{x-5}=1 \times 2^0$
$\Rightarrow x-5=0$
$\Rightarrow x=5$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
A briefcase was sold at a profit of $15\%$. If its cost price was $5\%$ less and it was sold for $Rs.35$ less, the gain would have been $20\%$. Find the cost price of the briefcase.
→Construct a trapezium $\text{ABCD}$ in which $AB = 4.6\ cm, BC = 6.4\ cm, CD = 5.6\ cm, \angle B = 60^\circ $ and $AD \| BC.$
→The ratio of passed and failed students in an examination was $3 : 1.$ Had $30$ less appeared and $10$ less failed, the ratio of passes to failures would have been $13 : 4.$ Find the number of students who appeared for the examination.
→Case Study : In 2015, the population of a town was 20000. During 2016 and 2017, it increases by 4% and 5% every year respectively. In 2018, due to an epidemic and migration to cities, the population decreases at 5%. Based on the above information answer the following questions:
1. The population of the town in 2016 was:
(a) 20800 (b) 20500 (c) 20460 (d) 20300
The difference in the population of the town at the end of the years 2017 and 2015 was:
(a) 1500 (b) 1700 (c) 1800 (d) 1840
The population of the town at the end of the year 2018 was:
(a) 20100 (b) 20500 (c) 20748 (d) 20850
Which of the following expressions gives the population at the end of the year 2018?
(a) $20000 \times \frac{104}{100} \times \frac{105}{100} \times \frac{105}{100}$
(b) $20000 \times \frac{104}{100} \times \frac{105}{100} \times \frac{95}{100}$
(c) $20000 \times \frac{104}{100} \times \frac{95}{100} \times \frac{95}{100}$
(d) $20000 \times \frac{96}{100} \times \frac{95}{100} \times \frac{95}{100}$
If the population of the town at the end of 2019 was 21163, then during 2019, the population increases at the rate of:
(a) 2% (b) 3% (c) 3.5% (d) 4%
→1. The population of the town in 2016 was:
(a) 20800 (b) 20500 (c) 20460 (d) 20300
The difference in the population of the town at the end of the years 2017 and 2015 was:
(a) 1500 (b) 1700 (c) 1800 (d) 1840
The population of the town at the end of the year 2018 was:
(a) 20100 (b) 20500 (c) 20748 (d) 20850
Which of the following expressions gives the population at the end of the year 2018?
(a) $20000 \times \frac{104}{100} \times \frac{105}{100} \times \frac{105}{100}$
(b) $20000 \times \frac{104}{100} \times \frac{105}{100} \times \frac{95}{100}$
(c) $20000 \times \frac{104}{100} \times \frac{95}{100} \times \frac{95}{100}$
(d) $20000 \times \frac{96}{100} \times \frac{95}{100} \times \frac{95}{100}$
If the population of the town at the end of 2019 was 21163, then during 2019, the population increases at the rate of:
(a) 2% (b) 3% (c) 3.5% (d) 4%
The sum of two numbers exceeds thrice the smaller by 2. If the difference between them is 19, find the numbers.
Find the amount and the compound interest payable annually on the following :$Rs.25000$ for $1 \frac{1}{2}$years at $10\%$ per annum.
→Draw a graph of the equation $2x + 3y + 5 = 0$, from the graph find the value of$:\ (i) x$, when $y = -3;(ii) y,$ when $x = 8$
→In the following figure, $O A=O C$ and $A B=B C$.
Prove that:$(i) \angle \mathrm{AOB}=900;(ii) \triangle \mathrm{AOD} \cong \triangle \mathrm{COD} ;(iii) A D=C D$
→Prove that:$(i) \angle \mathrm{AOB}=900;(ii) \triangle \mathrm{AOD} \cong \triangle \mathrm{COD} ;(iii) A D=C D$
A man lends $Rs. 12,500$ at $12\ \%\ $ for the first year, at $15\ \%\ $ for the second year and at $18\ \%\ $ for the third year. If the rates of interest are compounded yearly ; find the difference between the $C.I.$ fo the first year and the compound interest for the third year.
→Find the mean and median of all the positive factors of $72.$
→