QUESTION IMAGE
Question
solve the following exponential equation. express the solution set in terms of natural logarithms or common logarithms. then, use a calculator to obtain a decimal approximation for the solution. \\( 7^x + 7^x - 12 = 0 \\) (a) the solution set expressed in terms of logarithms is \\( \boxed{} \\). (use a comma to separate answers as needed. simplify your answer. use integers or fractions for any numbers in the expression. use \\( \ln \\) for natural logarithm and \\( \log \\) for common logarithm.) now use a calculator to obtain a decimal approximation for the solution. the solution set is \\( \boxed{} \\). (use a comma to separate answers as needed. round to two decimal places as needed.)
Step1: Isolate the exponential term
Given the equation \( 7^x + 7^x - 12 = 0 \), first combine like terms. Wait, actually, maybe it's a typo and the equation is \( 7^x + 7^x - 12 = 0 \)? Wait, no, maybe it's \( 7^x + x - 12 = 0 \)? Wait, the user's image shows \( 7^x + 7^x - 12 = 0 \)? Wait, no, looking at the image, the equation is \( 7^x + 7^x - 12 = 0 \)? Wait, no, maybe it's \( 7^x + x - 12 = 0 \)? Wait, the original equation in the image: "Solve the following exponential equation. Express the solution set in terms of natural logarithms or common logarithms. Then, use a calculator to obtain a decimal approximation for the solution. \( 7^x + 7^x - 12 = 0 \)" Wait, combining like terms: \( 2 \cdot 7^x - 12 = 0 \). Then, add 12 to both sides: \( 2 \cdot 7^x = 12 \). Divide both sides by 2: \( 7^x = 6 \).
Step2: Take logarithm on both sides
To solve for \( x \), take the natural logarithm (or common logarithm) of both sides. Let's use natural logarithm. So, \( \ln(7^x) = \ln(6) \). Using the logarithm power rule \( \ln(a^b) = b\ln(a) \), we get \( x\ln(7) = \ln(6) \). Then, solve for \( x \): \( x = \frac{\ln(6)}{\ln(7)} \).
Step3: Calculate the decimal approximation
Using a calculator, \( \ln(6) \approx 1.7918 \) and \( \ln(7) \approx 1.9459 \). Then, \( x \approx \frac{1.7918}{1.9459} \approx 0.92 \) (rounded to two decimal places).
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In terms of logarithms: \( x = \frac{\ln(6)}{\ln(7)} \) (or \( x = \log_7(6) \) using change of base formula). Decimal approximation: \( x \approx 0.92 \)