Question

9^log(10)*x+x^log(10)*9=6​

Answer 1

$9^{\log_{10}x}+x^{\log_{10}9}=6$

По свойству $a^{\log_{b}c}=c^{\log_ba}$ имеем:

$9^{\log_{10}x}+9^{\log_{10}x}=6;\\\\2\cdot9^{\log_{10}x}=6;\\\\9^{\log_{10}x}=3;\\\\3^{2\log_{10}x}=3^1;\\\\2\log_{10}x=1;\\\\\log_{10}x=0.5;\\\\x=10^{0.5};\\\\x=\sqrt{10}.$

ОТВЕТ: $\sqrt{10}$

Answer 2

To solve the equation $9^{\log_{10}(x)} \cdot x + x^{\log_{10}(9)} \cdot 9 = 6$, we can start by simplifying the exponents.

First, observe that $9^{\log_{10}(x)}$ can be written as $x^{\log_{10}(9)}$ because of the property $a^{\log_a(b)} = b$. So, the equation becomes:

$x^{\log_{10}(9)} \cdot x + x^{\log_{10}(9)} \cdot 9 = 6$

Now, factor out $x^{\log_{10}(9)}$:

$x^{\log_{10}(9)} \cdot (x + 9) = 6$

Now, set each factor equal to zero:

1. $x^{\log_{10}(9)} = 0$ (which is not possible since any positive number raised to any power will never be zero).
2. $x + 9 = 0$

Solve the second equation:

$x = -9$

However, we need to check if $x = -9$ is a valid solution. The original equation involves logarithms, and logarithms are only defined for positive numbers. Therefore, $x = -9$ is not a valid solution.

Hence, the given equation has no real solutions.