Question

Log^5x-log^5 35 < log^5 1/7

Answer 1

Log₅x-log₅35<log₅(1/17)
log₅x<log₅35+log₅(1/17)
log₅x<log₅(35*(1/17))
log₅x<log₅5
основание логарифма а=5, 5>1. знак неравенства не меняем
$\left \{ {{x\ \textgreater \ 0} \atop {x\ \textless \ 5}} \right.$

x∈(0;5)

Answer 2

5/35=36/х
5х=1260
х= 252

Answer 3

To solve the inequality:

log^5(x) - log^5(35) < log^5(1/7)

You can use the properties of logarithms to simplify and solve it. Here's the step-by-step process:

1. Use the property of logarithms that states log^a(b) - log^a(c) = log^a(b/c):

log^5(x/35) < log^5(1/7)

2. Now, we have an inequality with logarithms of the same base (base 5) on both sides. To eliminate the logarithms, we can exponentiate both sides with base 5:

5^(log^5(x/35)) < 5^(log^5(1/7))

3. By doing this, the left side simplifies to x/35, and the right side simplifies to 1/7:

x/35 < 1/7

4. To solve for x, multiply both sides by 35:

x < 5

So, the solution to the inequality is:

x < 5