Question

1. log_{5} (x+4)= 3log(5) 2+ log (5) 2

Answer 1

Log₅(x + 4) = 3log₅2  + log₅2
a) сначала ОДЗ
х + 4 > 0, ⇒ x > - 4
б) теперь решаем:
х + 4 = 2³*2
х + 4 = 16
х = 12 ( входит в ОДЗ)
в) Ответ: 12

Answer 2

Дано:
$log5(x+4)=3log(5)2+log(5)2$

Решение:
$log5(x+4)=3log(5)2+log(5)2$
$ODZ=x+4>0$
$ODZ=x>-4$
$x+4=2^{3}*2$
$2^{3}*2=2^{3}*2^{1}=2^{3+1}=2^{4}=16$
$x+4=16$
$x=16-4$
$x=12$

Ответ:
$12$

Answer 3

To solve the equation $\log_{5} (x+4) = 3\log_{5} 2 + \log_{5} 2$, we can use the properties of logarithms to simplify and solve for $x$:

Step 1: Apply the logarithmic properties First, we'll use the properties of logarithms to simplify the equation: $\log_{5} (x+4) = \log_{5} (2^3 \cdot 2)$

Step 2: Simplify the right side $\log_{5} (x+4) = \log_{5} (8 \cdot 2)$ $\log_{5} (x+4) = \log_{5} 16$

Step 3: Remove the logarithm Since both sides of the equation have the same base (base 5 logarithm), we can remove the logarithm and equate the expressions inside the logarithms: $x + 4 = 16$

Step 4: Solve for $x$ Subtract 4 from both sides: $x = 16 - 4$ $x = 12$

So, the solution to the equation is $x = 12$.