log(4)(3x + 4)^5 = 15
x = 20? And is this the only solution?
(3x+4)^5 = 4^15 = (4^3)^5
So,
3x+4 = 4^3 = 64
x = 20
That's the only solution I see.
To solve the equation log₄(3x + 4)^5 = 15, we first need to isolate the logarithm expression. Here are the step-by-step instructions:
Step 1: Rewrite the equation in exponential form.
Since log₄(3x + 4)^5 = 15, we can rewrite it as 4^15 = (3x + 4)^5.
Step 2: Take the fifth root of both sides.
∛(4^15) = ∛((3x + 4)^5).
Step 3: Simplify.
4^3 = 3x + 4.
Step 4: Solve for x.
64 = 3x + 4.
60 = 3x.
x = 20.
According to these steps, x = 20 is the solution to the equation. However, to determine if this is the only solution, we need to check for extraneous solutions.
To do this, we substitute the value of x back into the original equation and see if it holds true.
Let's plug x = 20 into the original equation:
log₄(3(20) + 4)^5 = 15.
log₄(64)^5 = 15.
log₄(2^6)^5 = 15.
log₄(2^30) = 15.
log₄(1073741824) = 15.
(log₄(4)) * (log₄(268435456)) = 15.
(1) * 9 = 15.
9 = 15.
Since 9 does not equal 15, we can conclude that x = 20 is not a valid solution. Therefore, there are no solutions to the equation log₄(3x + 4)^5 = 15.