Question

Loq0,7(x^-4x-5)=loq0,7(7-3x)

Answer 1

X^2 - 4x - 5 = 7 - 3x
x^2 - x - 12 = 0
x1= 4; x2 = -3

Answer 2

It seems like you have an equation involving logarithms that you'd like to solve. To simplify the notation, I assume you meant "log" when you wrote "loq."

The equation you provided is: $\log_{0.7}(x^{-4x-5}) = \log_{0.7}(7-3x)$

To solve this equation, we can use the property of logarithms that states: $\log_b(a) = \log_b(c) \implies a = c$

Using this property, we can equate the expressions inside the logarithms: $x^{-4x-5} = 7 - 3x$

Now, let's solve for "x."

Step 1: Get rid of the logarithm by converting both sides to the base of 0.7.

Step 2: Solve for "x."

Step 1: $x^{-4x-5} = 7 - 3x$

First, rewrite the right side in terms of the base 0.7: $\log_{0.7}(7 - 3x) = \log_{0.7}(0.7^2 \cdot (7 - 3x))$ $\log_{0.7}(7 - 3x) = \log_{0.7}(4.9 - 2.1x)$

Now the equation becomes: $\log_{0.7}(x^{-4x-5}) = \log_{0.7}(4.9 - 2.1x)$

Step 2: Since the bases of the logarithms are the same, we can equate the expressions inside the logarithms: $x^{-4x-5} = 4.9 - 2.1x$

To continue solving for "x," we need to get rid of the exponent on the left side. One way to do this is to take the logarithm of both sides (in any base, but we will choose 0.7 for simplicity):

$\log_{0.7}(x^{-4x-5}) = \log_{0.7}(4.9 - 2.1x)$

$(-4x - 5) \cdot \log_{0.7}(x) = \log_{0.7}(4.9 - 2.1x)$

Now, let's isolate the logarithm term on one side:

$\log_{0.7}(x) = \frac{\log_{0.7}(4.9 - 2.1x)}{-4x - 5}$

To solve for "x," we need to get rid of the logarithm. We can do this by converting the equation to an exponential form:

$x = 0.7^{\left(\frac{\log_{0.7}(4.9 - 2.1x)}{-4x - 5}\right)}$

Solving this equation for "x" will require numerical methods or graphing calculators as it involves both exponential and logarithmic functions. It might not have a simple algebraic solution.