
How do you solve \[3{{\left( 2 \right)}^{x-2}}+1=100\]?
Answer
549.9k+ views
Hint: We solve this using basic arithmetic and logarithm concepts. First we will group like terms together and then we will solve the equation. We will apply the logarithms concept needed while solving the equation we get. Then by simplifying we will get the solution.
Complete step by step solution:
Given equation
\[3{{\left( 2 \right)}^{x-2}}+1=100\]
Now we will subtract 1 from both sides of the equation.
After subtracting we will get
\[\Rightarrow 3{{\left( 2 \right)}^{x-2}}+1-1=100-1\]
By simplifying we will get
\[\Rightarrow 3{{\left( 2 \right)}^{x-2}}=99\]
Now we will divide the equation with 3 on both sides of the equation.
\[\Rightarrow \dfrac{3{{\left( 2 \right)}^{x-2}}}{3}=\dfrac{99}{3}\]
By simplifying we will get
\[\Rightarrow {{2}^{x-2}}=33\]
Here we have a formula that
If \[{{2}^{a}}=b\]then \[a={{\log }_{2}}b\]
Using the above formula we can write our equation as
\[\Rightarrow x-2={{\log }_{2}}33\]
Now to get the value of x we will add 2 on both sides of the equation.
By adding 2 we will get
\[\Rightarrow x-2+2={{\log }_{2}}33+2\]
By simplifying it we will get
\[\Rightarrow x={{\log }_{2}}33+2\]
We cannot simplify it further.
So by solving the equation we will get \[x={{\log }_{2}}33+2\].
This solution can also be written in different forms like
In decimal form
\[x=7.04439411\]
Using natural logarithms we will get
\[x=2+\dfrac{\ln \left( 33 \right)}{\ln \left( 2 \right)}\]
So any form of the answer is a correct solution.
Note: We can also do this in different ways. One way of doing this is by applying natural logarithms on both sides of the equation. First we will apply natural logs on both sides. After that by using various simplification techniques like expanding and distributive properties, taking out common terms etc.. we can arrive at the solution. We will get the solution in the form \[x=2+\dfrac{\ln \left( 33 \right)}{\ln \left( 2 \right)}\].
Complete step by step solution:
Given equation
\[3{{\left( 2 \right)}^{x-2}}+1=100\]
Now we will subtract 1 from both sides of the equation.
After subtracting we will get
\[\Rightarrow 3{{\left( 2 \right)}^{x-2}}+1-1=100-1\]
By simplifying we will get
\[\Rightarrow 3{{\left( 2 \right)}^{x-2}}=99\]
Now we will divide the equation with 3 on both sides of the equation.
\[\Rightarrow \dfrac{3{{\left( 2 \right)}^{x-2}}}{3}=\dfrac{99}{3}\]
By simplifying we will get
\[\Rightarrow {{2}^{x-2}}=33\]
Here we have a formula that
If \[{{2}^{a}}=b\]then \[a={{\log }_{2}}b\]
Using the above formula we can write our equation as
\[\Rightarrow x-2={{\log }_{2}}33\]
Now to get the value of x we will add 2 on both sides of the equation.
By adding 2 we will get
\[\Rightarrow x-2+2={{\log }_{2}}33+2\]
By simplifying it we will get
\[\Rightarrow x={{\log }_{2}}33+2\]
We cannot simplify it further.
So by solving the equation we will get \[x={{\log }_{2}}33+2\].
This solution can also be written in different forms like
In decimal form
\[x=7.04439411\]
Using natural logarithms we will get
\[x=2+\dfrac{\ln \left( 33 \right)}{\ln \left( 2 \right)}\]
So any form of the answer is a correct solution.
Note: We can also do this in different ways. One way of doing this is by applying natural logarithms on both sides of the equation. First we will apply natural logs on both sides. After that by using various simplification techniques like expanding and distributive properties, taking out common terms etc.. we can arrive at the solution. We will get the solution in the form \[x=2+\dfrac{\ln \left( 33 \right)}{\ln \left( 2 \right)}\].
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

10 examples of friction in our daily life

