4^x-3=6 please help

Mathematics

1 answer:

SashulF [63]3 years ago

3 0

$4^{x+3}=6\ \ \ \ |\log_4\\\\\log_44^{x+3}=\log_46\ \ \ |\text{Use}\ \log_ab^n=n\log_ab\\\\(x+3)\log_44=\log_46\ \ \ \ |\text{Use}\ \log_aa=1\\\\x+3=\log_46\ \ \ |-3\\\\x=\log_46-3\ \ \ \ |\text{Use}\ b=\log_aa^b\\\\x=\log_46-\log_44^3\\\\x=\log_46-\log_464\ \ \ |\text{Use}\ \log_ab-\log_ac=\log_a\dfrac{b}{c}\\\\x=\log_4\dfrac{6}{64}\\\\x=\log_4\dfrac{3}{32}$

You might be interested in

What is the area of a circle whose equation is (x-3)² + (y + 1) ² =81?

kupik [55]

Okay, first, given the equation, we need to find out what the radius of the circle is. Let us state the general equation of a circle:

$\frac{(x-x_{1})^2}{r^2}+ \frac{(y-y_{1})_^2}{r^2}=1$

Where $(x_{1}, y_{1})$ is the centre of the circle. In this case, we don't need to know the centre. Just the radius.

Let us start by converting the equation into standard for, which I typed above. Divide both sides by 81.

$\frac{(x-3)^2}{81}+ \frac{(y+1)^2}{81}=1$

Great! We now know the radius of the circle. It is $\sqrt{81}$ because it is the bottom fraction. Now we know that the radius is 9.

So now lets input this into the area of circle formula:

$A=πr^2$

Now we insert our radius.

$A=9^2π$

$=A=81π$

You can convert that into a decimal if you wish.

Hope this helped!

~Cam943, Moderator

4 0

3 years ago

Refer to the photo :D

OlgaM077 [116]

well, we know the ceiling is 6+2/3 high, and Eduardo has 4+1/2 yards only, how much more does he need, well, is simply their difference, let's firstly convert the mixed fractions to improper fractions and then subtract.

$\stackrel{mixed}{6\frac{2}{3}}\implies \cfrac{6\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{20}{3}} ~\hfill \stackrel{mixed}{4\frac{1}{2}}\implies \cfrac{4\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{9}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{20}{3}-\cfrac{9}{2}\implies \stackrel{using ~~\stackrel{LCD}{6}}{\cfrac{(2\cdot 20)-(3\cdot 9)}{6}}\implies \cfrac{40-27}{6}\implies \cfrac{13}{6}\implies\blacktriangleright 2\frac{1}{6} \blacktriangleleft$