Question

Log7(x) - 1=6 logx(7)

Answer 1

Доп.условие x>0
$log_7x-1=\frac{6}{log_7x}\\log_7^2x-log_7x-6=0\\log_7x=-2,log_7x=3\\log_7x=log_77^{-2},log_7x=log_77^3\\x=\frac{1}{49};x=343$
Ответ: $\frac{1}{49};x=343$

Answer 2

To solve the equation Log7(x) - 1=6 logx(7), we can use the change of base formula to express both logarithms in terms of a common base. Let's use base 10:

Log7(x) = Log10(x) / Log10(7)

logx(7) = log10(7) / log10(x)

Substituting these expressions into the original equation, we get:

Log10(x) / Log10(7) - 1 = 6 log10(7) / log10(x)

Multiplying both sides by Log10(7) log10(x) to eliminate the denominators, we get:

Log10(x)^2 - Log10(7) Log10(x) - 6 Log10(7) = 0

This is a quadratic equation in Log10(x). We can solve it using the quadratic formula:

Log10(x) = [Log10(7) ± sqrt(Log10(7)^2 + 24 Log10(7))] / 2

Log10(x) = [Log10(7) ± Log10(7)(sqrt(25))] / 2

Log10(x) = [Log10(7) ± 5 Log10(7)] / 2

Simplifying each solution, we get:

Log10(x) = 3 Log10(7) or Log10(x) = -2 Log10(7)

Using the definition of logarithms, we can rewrite these solutions as:

x = 7^3 or x = 7^(-2)

Therefore, the solutions of the equation Log7(x) - 1=6 logx(7) are x = 343 or x = 1/49.