How do you solve \[{{5}^{x+1}}=125\]?
Answer
624.3k+ views
Hint: This type of problem can be solved by converting the base to a common term and then equating the powers with the variable. First, we have to consider the right-hand side of the given equation. Write 125 as the cube of 5. Now the bases to the LHS and RHS are the same. Using the rule ‘if\[{{a}^{n}}={{a}^{m}}\], then n=m’, we get x+1=3. Subtract 1 from both the sides of the expression and do necessary calculations to get the value of x which is the required answer.
Complete step-by-step solution:
According to the question, we are asked to solve \[{{5}^{x+1}}=125\].
We have been given the equation is \[{{5}^{x+1}}=125\]. ---------(1)
Let us first consider the right-hand side of the given equation (1).
RHS=125
We know that \[125=5\times 5\times 5\].
Therefore, we can write 125 as the cube of 5 that is \[{{5}^{3}}\].
Therefore, RHS=\[{{5}^{3}}\].
Substitute the RHS in equation (1).
\[\Rightarrow {{5}^{x+1}}={{5}^{3}}\]
We find that the LHS and RHS have the same term 5 with the power difference.
We know that, if \[{{a}^{n}}={{a}^{m}}\], then n=m.
Using this rule in the above expression, we get
$x+1=3$
Subtract 1 from both the sides of the expression, we get
$x+1-1=3-1$
We know that terms with the same magnitude and opposite signs cancel out.
Therefore, we get
$x=3-1$
On further simplification, we get
$x=2$
Hence, the value of x in \[{{5}^{x+1}}=125\] is 2.
Note: Whenever we get such types of problems, we should always compare the right-hand side of the equation with the left-hand side of the equation. Convert the LHS and RHS of the equation in such a way that we can equate their powers. Avoid calculation mistakes based on sign conventions. Do not neglect the negative sign in the power.
Complete step-by-step solution:
According to the question, we are asked to solve \[{{5}^{x+1}}=125\].
We have been given the equation is \[{{5}^{x+1}}=125\]. ---------(1)
Let us first consider the right-hand side of the given equation (1).
RHS=125
We know that \[125=5\times 5\times 5\].
Therefore, we can write 125 as the cube of 5 that is \[{{5}^{3}}\].
Therefore, RHS=\[{{5}^{3}}\].
Substitute the RHS in equation (1).
\[\Rightarrow {{5}^{x+1}}={{5}^{3}}\]
We find that the LHS and RHS have the same term 5 with the power difference.
We know that, if \[{{a}^{n}}={{a}^{m}}\], then n=m.
Using this rule in the above expression, we get
$x+1=3$
Subtract 1 from both the sides of the expression, we get
$x+1-1=3-1$
We know that terms with the same magnitude and opposite signs cancel out.
Therefore, we get
$x=3-1$
On further simplification, we get
$x=2$
Hence, the value of x in \[{{5}^{x+1}}=125\] is 2.
Note: Whenever we get such types of problems, we should always compare the right-hand side of the equation with the left-hand side of the equation. Convert the LHS and RHS of the equation in such a way that we can equate their powers. Avoid calculation mistakes based on sign conventions. Do not neglect the negative sign in the power.
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