Question

1. 4^(х+3)/х + 4^(2х+3)/х = 320 2. 0,8^(0,8-5х) = 0,5√5

Answer 1

$3)0,8^{0,8-5x} =0,5\sqrt{5}\\\\(\frac{4}{5})^{0,8-5x}=\sqrt{\frac{1}{4} *5}\\\\(\frac{4}{5} )^{0,8-5x}=(\frac{5}{4})^{\frac{1}{2} }\\\\(\frac{5}{4})^{5x-0,8}=(\frac{5}{4})^{0,5}\\\\5x-0,8=0,5\\\\5x=1,3\\\\x=0,26\\\\4)4^{\frac{x+3}{x} }+4^{\frac{2x+3}{x} } =320\\\\4^{1}*4^{\frac{3}{x} }+4^{2}*4^{\frac{3}{x}}=320\\\\4*4^{\frac{3}{x} }+16*4^{\frac{3}{x} } =320\\\\20*4^{\frac{3}{x} }=320\\\\4^{\frac{3}{x} }=16\\\\4^{\frac{3}{x} }=4^{2}\\\\\frac{3}{x}=2\\\\x=1,5$

Answer 2

Let's solve these equations step by step:

1. $4^{(x+3)/x} + 4^{(2x+3)/x} = 320$

First, let's simplify the equation. We can rewrite $4^{(x+3)/x}$ and $4^{(2x+3)/x}$ as follows:

$4^{(x+3)/x} = 4^{3/x}$ $4^{(2x+3)/x} = 4^{3(2/x)} = 4^{6/x}$

Now, the equation becomes:

$4^{3/x} + 4^{6/x} = 320$

Let's make a substitution to simplify further. Let $u = 4^{3/x}$, then the equation becomes:

$u + u^2 = 320$

Now it's a quadratic equation in terms of u. Rearrange it:

$u^2 + u - 320 = 0$

This is a quadratic equation, and you can solve it using the quadratic formula:

$u = (-b ± √(b² - 4ac)) / (2a)$

Where a = 1, b = 1, and c = -320 in our case.

$u = (-1 ± √(1² - 4(1)(-320))) / (2(1))$

$u = (-1 ± √(1 + 1280)) / 2$

$u = (-1 ± √1281) / 2$

Now, we have two possible solutions for u:

1. $u_1 = (-1 + √1281) / 2$
2. $u_2 = (-1 - √1281) / 2$

Now, recall that we made the substitution $u = 4^{3/x}$. We can use these values of u to find x:

1. $4^{3/x} = (-1 + √1281) / 2$
2. $4^{3/x} = (-1 - √1281) / 2$

Now, isolate x:

For the first equation:

$3/x = log_4((-1 + √1281) / 2)$

And for the second equation:

$3/x = log_4((-1 - √1281) / 2)$

You can solve for x by taking the reciprocal of both sides after finding the logarithm.

2. $0.8^{(0.8-5x)} = 0.5√5$

First, let's isolate the exponent:

$0.8^{(0.8-5x)} = 0.5√5$

Now, take the natural logarithm (ln) of both sides to solve for x:

$ln(0.8^{(0.8-5x)}) = ln(0.5√5)$

Using the properties of logarithms, we can bring the exponent down:

$(0.8-5x)ln(0.8) = ln(0.5√5)$

Now, solve for x:

$0.8 - 5x = ln(0.5√5) / ln(0.8)$

$5x = 0.8 - (ln(0.5√5) / ln(0.8))$

$x = (0.8 - (ln(0.5√5) / ln(0.8))) / 5$

You can calculate the numerical value of x using a calculator.