what is the true solution to 3 in 2 + in 8 = 2 in (4x)
Accepted Solution
A:
The first step for solving this equation is to determine the defined range. 3㏑(2) + ㏑(8) = 2㏑(4x), x > 0 Write the number 8 in exponential form. 3㏑(2) + ㏑(2³) = 2㏑(4x) Using ㏑([tex] a^{x} [/tex]) = x × ㏑(a),, transform the expression. 3㏑(2) + 3㏑(2) = 2㏑(4x) Now collect the like terms on the left side of the equation. 6㏑(2) = 2㏑(4x) Switch the sides of the equation. 2㏑(4x) = 6㏑(2) Using x × ㏑(a) = ㏑([tex] a^{x} [/tex]),, transform the expression on the left side of the equation. ㏑((4x)²) = 6㏑(2) Using x × ㏑(a) = ㏑([tex] a^{x} [/tex]),, transform the expression on the right side of the equation. ㏑((4x)²) = ㏑([tex] 2^{6} [/tex]) Since the bases of the logarithms are the same,, you need to set the arguments equal. (4x)² = [tex] 2^{6} [/tex] Take the square root of both sides of the equation and remember to use both the positive and negative roots. 4x = +/- 8 Now separate the equation into 2 possible cases. 4x = 8 4x = -8 Solve the top equation for x. x = 2 Solve the bottom equation for x. x = 2 , x > 0 x = -2 Lastly,, check if the solution is in the defined range to find your final answer. x = 2 This means that the correct answer to your question is x = 2. Let me know if you have any further questions :)