How do you solve ${8^{x + 1}} = 16$ ?
Answer
616.2k+ views
Hint: In this question, we need to find the solution for an exponential equation. Here we try to make the bases same on both sides of the equation by some basic calculations. Firstly, we replace 8 by a cube of 2, i.e. ${2^3}$ and then 16 by 2 to the power 4, i.e. ${2^4}$. After making this, we get the same base on both sides which is 2. Then we use the fact that, if the bases are the same then the equation is equal only if the exponents are also equal and find the required solution.
Complete step-by-step solution:
Given an equation of the form,
${8^{x + 1}} = 16$..............…… (1)
We are asked to solve the exponential equation given in the equation (1).
Note that we approach the above exponential equation, by making the bases same on both sides.
Firstly, we write the numbers on both sides of the equation by power of 2.
We know that 8 is equal to 2 to the power three.
Hence we write 8 as, $8 = {2^3}$.
Also we know that 16 is equal to 2 to the power four.
Hence we can write 16 as, $16 = {2^4}$.
So we replace 8 and 16 in the equation (1) we get,
$ \Rightarrow {\left( {{2^3}} \right)^{x + 1}} = {2^4}$
This can also be written as,
$ \Rightarrow {\left( 2 \right)^{3(x + 1)}} = {2^4}$
Now the base on both sides of the equation is the same which is 2.
We have the result that if the expression having the same bases is equal only if the exponents or indices are also equal.
Hence equating the indices, we get,
$ \Rightarrow 3(x + 1) = 4$
Now multiplying 3 inside the parenthesis to each term we get,
$ \Rightarrow 3 \cdot x + 3 \cdot 1 = 4$
$ \Rightarrow 3x + 3 = 4$
Move all the terms which do not contain the x term to the R.H.S.
Transferring 3 to the other side we get,
$ \Rightarrow 3x = 4 - 3$
$ \Rightarrow 3x = 1$
Now divide throughout by 3 we get,
$ \Rightarrow \dfrac{{3x}}{3} = \dfrac{1}{3}$
$ \Rightarrow x = \dfrac{1}{3}$
Hence the solution for the equation ${8^{x + 1}} = 16$ is $x = \dfrac{1}{3}$.
Note: We can verify the obtained answer is correct by substituting back it in the given equation. If the equation satisfies, i.e. if we get L.H.S. is equal to R.H.S. then the obtained value is correct. Otherwise our answer is wrong.
Students must remember the rules of exponents to simplify such problems. We need to be careful while applying the rules. It is necessary to use the correct rule to split the terms and simplify the answer.
The rules of exponents are given below.
(1) Multiplication rule : ${a^m} \cdot {a^n} = {a^{m + n}}$
(2) Division rule : $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$
(3) Power of a power rule : ${({a^m})^n} = {a^{mn}}$
(4) Power of a product rule : ${(ab)^m} = {a^m}{b^m}$
(5) Power of a fraction rule : ${\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}$
(6) Zero exponent : ${a^0} = 1$
(7) Negative exponent : ${a^{ - x}} = \dfrac{1}{{{a^x}}}$
Complete step-by-step solution:
Given an equation of the form,
${8^{x + 1}} = 16$..............…… (1)
We are asked to solve the exponential equation given in the equation (1).
Note that we approach the above exponential equation, by making the bases same on both sides.
Firstly, we write the numbers on both sides of the equation by power of 2.
We know that 8 is equal to 2 to the power three.
Hence we write 8 as, $8 = {2^3}$.
Also we know that 16 is equal to 2 to the power four.
Hence we can write 16 as, $16 = {2^4}$.
So we replace 8 and 16 in the equation (1) we get,
$ \Rightarrow {\left( {{2^3}} \right)^{x + 1}} = {2^4}$
This can also be written as,
$ \Rightarrow {\left( 2 \right)^{3(x + 1)}} = {2^4}$
Now the base on both sides of the equation is the same which is 2.
We have the result that if the expression having the same bases is equal only if the exponents or indices are also equal.
Hence equating the indices, we get,
$ \Rightarrow 3(x + 1) = 4$
Now multiplying 3 inside the parenthesis to each term we get,
$ \Rightarrow 3 \cdot x + 3 \cdot 1 = 4$
$ \Rightarrow 3x + 3 = 4$
Move all the terms which do not contain the x term to the R.H.S.
Transferring 3 to the other side we get,
$ \Rightarrow 3x = 4 - 3$
$ \Rightarrow 3x = 1$
Now divide throughout by 3 we get,
$ \Rightarrow \dfrac{{3x}}{3} = \dfrac{1}{3}$
$ \Rightarrow x = \dfrac{1}{3}$
Hence the solution for the equation ${8^{x + 1}} = 16$ is $x = \dfrac{1}{3}$.
Note: We can verify the obtained answer is correct by substituting back it in the given equation. If the equation satisfies, i.e. if we get L.H.S. is equal to R.H.S. then the obtained value is correct. Otherwise our answer is wrong.
Students must remember the rules of exponents to simplify such problems. We need to be careful while applying the rules. It is necessary to use the correct rule to split the terms and simplify the answer.
The rules of exponents are given below.
(1) Multiplication rule : ${a^m} \cdot {a^n} = {a^{m + n}}$
(2) Division rule : $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$
(3) Power of a power rule : ${({a^m})^n} = {a^{mn}}$
(4) Power of a product rule : ${(ab)^m} = {a^m}{b^m}$
(5) Power of a fraction rule : ${\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}$
(6) Zero exponent : ${a^0} = 1$
(7) Negative exponent : ${a^{ - x}} = \dfrac{1}{{{a^x}}}$
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