Question

Log3(5^x-1)+Log3(5^x+1)=1+Log3(2)

Answer 1

Вот решение на листке

Answer 2

To solve the equation $\log_3(5^x-1)+\log_3(5^x+1)=1+\log_3(2)$, we can use logarithmic properties to simplify it.

Recall the logarithmic property: $\log_b(a) + \log_b(c) = \log_b(ac)$.

Using this property, we can rewrite the equation as follows:

$\log_3((5^x-1)(5^x+1)) = \log_3(2) + 1$.

Now, another logarithmic property states that $\log_b(a) + c = \log_b(a \cdot b^c)$. Using this property, we get:

$\log_3((5^x-1)(5^x+1)) = \log_3(2 \cdot 3)$.

Since the bases of both logarithms are the same, the arguments must also be equal. Therefore, we have:

$(5^x-1)(5^x+1) = 2 \cdot 3$.

Now, let's solve for $x$:

$5^x \cdot 5^x - 1 = 6$.

$(5^x)^2 - 1 = 6$.

$(5^x)^2 = 7$.

Now, take the square root of both sides:

$5^x = \sqrt{7}$.

Finally, solve for $x$ by taking the logarithm base 5 on both sides:

$x = \log_5(\sqrt{7})$.

Using the property $\log_b(\sqrt{a}) = \frac{1}{2}\log_b(a)$, we get:

$x = \frac{1}{2}\log_5(7)$.

This is the exact solution for $x$ in terms of logarithms. If you need a numerical approximation, you can calculate $\log_5(7)$ using a calculator and then divide it by 2 to find the value of $x$.