Solve the logarithmic equation: log 6 + x log 4 = log 4 + log (32 + 4^x)

Question

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Step 1: Apply the logarithm power rule, a log b = log (b^a), to the term x log 4. log 6 + log (4^x) = log 4 + log (32 + 4^x); Step 2: Apply the logarithm product rule, log a + log b = log (ab), to both sides of the equation. log (6 · 4^x) = log (4 · (32 + 4^x)) Step 3: Since the logarithms on both sides are equal, their arguments must be equal. 6 · 4^x = 4 · (32 + 4^x); Step 4: Let y = 4^x to simplify the equation and solve for y. 6y = 4(32 + y); 6y = 128 + 4y Subtract 4y from both sides: 6y − 4y = 128; 2y = 128 Divide by 2: y = 128/2; y = 64; Step 5: Substitute y = 4^x back into the equation and solve for x. 4^x = 64; Recognize that 64 can be expressed as a power of 4: 64 = 4³. 4^x = 4³ Since the bases are the same, the exponents must be equal. x = 3 The final answer is 3.

Solve the logarithmic equation: log 6 + x log 4 = log 4 + log (32 + 4^x)

Answer

Step-by-step solution

Apply the logarithm power rule, $a \log b = \log (b^a)$ , to the term $x \log 4$ .

Apply the logarithm product rule, $\log a + \log b = \log (ab)$ , to both sides of the equation.

Since the logarithms on both sides are equal, their arguments must be equal.

Let $y = 4^x$ to simplify the equation and solve for $y$ .

Substitute $y = 4^x$ back into the equation and solve for $x$ .

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Solve the logarithmic equation: log 6 + x log 4 = log 4 + log (32 + 4^x)

Answer

Step-by-step solution

Apply the logarithm power rule, alog⁡b=log⁡(ba)a \log b = \log (b^a)alogb=log(ba), to the term xlog⁡4x \log 4xlog4.

Apply the logarithm product rule, log⁡a+log⁡b=log⁡(ab)\log a + \log b = \log (ab)loga+logb=log(ab), to both sides of the equation.

Since the logarithms on both sides are equal, their arguments must be equal.

Let y=4xy = 4^xy=4x to simplify the equation and solve for yyy.

Substitute y=4xy = 4^xy=4x back into the equation and solve for xxx.

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More Mathematics Questions

Apply the logarithm power rule, $a \log b = \log (b^a)$ , to the term $x \log 4$ .

Apply the logarithm product rule, $\log a + \log b = \log (ab)$ , to both sides of the equation.

Let $y = 4^x$ to simplify the equation and solve for $y$ .

Substitute $y = 4^x$ back into the equation and solve for $x$ .