Problem 93
Question
Solve each equation. $$ 5^{2 x} \cdot 5^{4 x}=125 $$
Step-by-Step Solution
Verified Answer
The solution to the equation is \(x = 0.5\).
1Step 1: Combine the powers with the same base
Since the bases are the same (both are 5), the rule of multiplication of powers with the same base can be used. The rule states that \(a^{m} * a^{n} = a^{m+n}\). Here our base is 5, therefore we have \(5^{2x} * 5^{4x} = 5^{2x + 4x} = 5^{6x}\).
2Step 2: Swap the left and right side of the equation
The equation now is \(5^{6x} = 125\). Swap the two sides for convenience. This doesn't affect the solution: \(125 = 5^{6x}\).
3Step 3: Transfer 125 to base 5
To apply the principles of logarithms later, it is helpful to have the number 125 in the base of 5. The number 125 is actually \(5^3\), therefore, the equation now is \(5^3 = 5^{6x}\).
4Step 4: Equate the powers of the equation
Now the bases are the same (i.e., base 5), we can equate the powers. This gives us the equation \(3 = 6x\).
5Step 5: Solve for x
Finally, we can solve for \(x\) by dividing both sides by 6. This gives us \(x = 3/6 = 0.5\).
Other exercises in this chapter
Problem 92
In Exercises 81–100, evaluate or simplify each expression without using a calculator. $$ \ln \frac{1}{e^{7}} $$
View solution Problem 92
Graph \(f(x)=2^{x}\) and its inverse function in the same rectangular coordinate system.
View solution Problem 93
In Exercises 81–100, evaluate or simplify each expression without using a calculator. $$ e^{\ln 125} $$
View solution Problem 93
The hyperbolic cosine and hyperbolic sine functions are defined by $$ \cosh x=\frac{e^{x}+e^{-x}}{2} \quad \text { and } \quad \sinh x=\frac{e^{x}-e^{-x}}{2} $$
View solution