Question

Log4^(2x+5)>log4^25 ; 2log5^25+3log2^64. помогите решить

Answer 1

В приложении ............

Answer 2

To solve the inequalities, we can use the properties of logarithms. Here are the steps to solve each inequality:

1. $\log_4^{(2x+5)} > \log_4^{25}$

Step 1: Apply the property of logarithms that states if $\log_a{b} > \log_a{c}$, then $b > c$.

Step 2: Set the exponents inside the logarithms equal to each other and solve for $x$:

$2x + 5 > 25$

Step 3: Now, solve for $x$:

$2x > 25 - 5$

$2x > 20$

$x > 10$

So the solution to the inequality is $x > 10$.

2. $2 \log_5^{25} + 3 \log_2^{64}$

Step 1: Simplify the logarithmic expressions.

$2 \log_5^{25} = 2 \cdot 2 = 4$

$3 \log_2^{64} = 3 \cdot 6 = 18$

Step 2: Now add the simplified expressions:

$4 + 18 = 22$

So, $2 \log_5^{25} + 3 \log_2^{64} = 22$.

There you have the solutions to both the inequalities and the expression.