Question

2^x-1≥8 log0.2(2x+3)≥-3 решите неравенства

Answer 1

$2^x - 1 \geq 8 \\ 2^x \geq 9 \\ 2^x \geq 2^{log_29} \\ x \geq log_29$

$2x + 3 \ \textgreater \ 0 \\ x\ \textgreater \ - \frac{3}{2} \\ log_{ \frac{1}{5} }(2x -3) \geq -3 \\ log_5(2x-3) \leq 3 \\ log_5(2x-3) \leq log_5125 \\ 2x + 3 \leq 125 \\ 2x \leq 122 \\ x \leq 61$

x ∈ $(- \frac{3}{2} ;61]$

Answer 2

This is a system of logarithmic inequalities. To solve it, we need to use the properties of logarithms and inequalities. Here are the steps:

1. Rewrite the first inequality as 2^x-1-8≥0 and the second inequality as log0.2(2x+3)+3≥0. 2. Solve the first inequality by using the exponent property of inequality: if a>0 and b>0, then a^x>b^x if and only if x>loga(b). In this case, we have 2^x-1-8≥0 if and only if x≥log2(9). So the solution set for the first inequality is [log2(9),∞). 3. Solve the second inequality by using the change of base formula: loga(b)=logc(b)/logc(a). In this case, we have log0.2(2x+3)+3≥0 if and only if log10(2x+3)/log10(0.2)+3≥0. Simplify the expression and move all the terms to one side: log10(2x+3)-3log10(0.2)≥-3. Apply the power property of logarithm: loga(b^c)=cloga(b). We get log10(2x+3)-log10(0.2^3)≥-3. Apply the quotient property of logarithm: loga(b/c)=loga(b)-loga(c). We get log10((2x+3)/(0.2^3))≥-3. Apply the inverse property of logarithm: loga(a^b)=b. We get (2x+3)/(0.2^3)≥10^-3. Simplify the expression and cross multiply: 2x+3≥0.008. Subtract 3 from both sides: 2x≥-2.992. Divide both sides by 2: x≥-1.496. So the solution set for the second inequality is [-1.496,∞). 4. Find the intersection of the solution sets for the two inequalities. This is the final solution for the system of inequalities. The intersection is [log2(9),∞), which means x≥log2(9).

You can check your answer by using an online calculator or by graphing the inequalities. I hope this helps you understand how to solve logarithmic inequalities.