What is the solution to the equation below

What is the solution to the equation below? 3log4⁡x=log4⁡32+log4⁡2 x = -8 x = -4 x = 4 x = 8
x = -8 x = -4 x = 4 x = 8,

The Correct Answer and Explanation is:

We are given the equation: 3log⁡4x=log⁡432+log⁡423 \log_4 x = \log_4 32 + \log_4 23log4​x=log4​32+log4​2

Step-by-Step Solution

Step 1: Use the property of logarithms

The right-hand side uses the property of logarithms: log⁡bA+log⁡bB=log⁡b(AB)\log_b A + \log_b B = \log_b (AB)logb​A+logb​B=logb​(AB)

So we combine the right-hand side: log⁡432+log⁡42=log⁡4(32×2)=log⁡464\log_4 32 + \log_4 2 = \log_4 (32 \times 2) = \log_4 64log4​32+log4​2=log4​(32×2)=log4​64

Now the equation becomes: 3log⁡4x=log⁡4643 \log_4 x = \log_4 643log4​x=log4​64

Step 2: Use the property of logarithms

Use the power property: alog⁡bx=log⁡b(xa)a \log_b x = \log_b (x^a)alogb​x=logb​(xa)

So: 3log⁡4x=log⁡4(x3)3 \log_4 x = \log_4 (x^3)3log4​x=log4​(x3)

Now the equation is: log⁡4(x3)=log⁡464\log_4 (x^3) = \log_4 64log4​(x3)=log4​64

Since the bases and logs are equal, we equate the arguments: x3=64x^3 = 64×3=64

Step 3: Solve the equation

Take the cube root of both sides: x=643=4x = \sqrt[3]{64} = 4x=364​=4

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Final Answer:

x = 4

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Explanation

This problem tests your understanding of logarithmic rules, particularly properties for combining and simplifying logarithmic expressions. The given equation includes a logarithm with a base of 4, and we need to simplify both sides to isolate the variable xxx.

First, we simplify the right-hand side using the product rule of logarithms: log⁡bA+log⁡bB=log⁡b(AB)\log_b A + \log_b B = \log_b (AB)logb​A+logb​B=logb​(AB). Applying this, the sum log⁡432+log⁡42\log_4 32 + \log_4 2log4​32+log4​2 becomes log⁡4(32⋅2)=log⁡464\log_4 (32 \cdot 2) = \log_4 64log4​(32⋅2)=log4​64. Recognizing powers of 2 helps here: 32=2532 = 2^532=25 and 2=212 = 2^12=21, so 64=2664 = 2^664=26, which is important for evaluating logarithms.

Next, on the left-hand side, we simplify 3log⁡4×3 \log_4 x3log4​x using the power rule of logarithms: alog⁡bx=log⁡b(xa)a \log_b x = \log_b (x^a)alogb​x=logb​(xa). Thus, 3log⁡4x=log⁡4(x3)3 \log_4 x = \log_4 (x^3)3log4​x=log4​(x3).

Now the equation becomes log⁡4(x3)=log⁡464\log_4 (x^3) = \log_4 64log4​(x3)=log4​64. Because the logs have the same base and the expressions are equal, their arguments must be equal too. That gives us the equation x3=64x^3 = 64×3=64.

To solve x3=64x^3 = 64×3=64, take the cube root of both sides. Since 64=4364 = 4^364=43, we find x=4x = 4x=4.

In conclusion, using logarithmic properties allowed us to transform and simplify the equation, ultimately solving it by equating arguments of the logs. The correct solution is:
x = 4.