32 x (8*) = 3 ; give an exact solution for * (* is a power)
any help would be appreciated
I don't know what class this is for, but this is a good basic algebra problem. I would stop using "x" to mean multiply because in algebra you'll use "x" to mean an unknown number most of the time. You can use the "*" as a symbol for multiplication, use parentheses like 8(4) to mean 8 times 4.
Let "x" be the number that we want to find that fits the problems guidelines. It doesn't matter what it is yet, but we know it is a real number, so just call it x.
So $\displaystyle 32*8*x=3$ meaning 32 times 8 times x equals 3. We want to rearrange this to get this in the form of x=... so let's do that. The opposite operation of multiplication is division, so if we want to "cancel" the 32, we can divide by 32 on both sides of the equation and still keep the = sign. Do the same to the 8 and you should get $\displaystyle x=\frac{3}{32*8}$
Hello, highsamurai!
Is that an exponent?
Solve for $\displaystyle x\!:\quad32\cdot8^x \:=\: 3$
We have: .$\displaystyle 2^5(2^3)^x \:=\:3 \quad\Rightarrow\quad 2^{3x+5} \:=\:3 $
Take logs: .$\displaystyle \ln\left(2^{3x+5}\right) \:=\:\ln(3) \quad\Rightarrow\quad(3x+5)\ln(2) \:=\:\ln(3)$
. . $\displaystyle 3x+5 \:=\:\frac{\ln(3)}{\ln(2)} \quad\Rightarrow\quad 3x \:=\:\frac{\ln(3)}{\ln(2)} - 5 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{1}{3}\left(\frac{\ln(3)}{\ln(2)} - 5\right)}$
thnx man that helped a lot.
I also have another one if you dont mind. I just dont know what to do next.
log10 (x - 5) + log10 (3x + 6) = log10 (x+4)
log10 (x - 5)(3x + 6) = log10 (x + 4)
(x - 5)(3x + 6) = (x + 4)
3x^2 - 9x - 30 = x + 4
3x^2 - 9x - 34 = x
(the ^2 is squared)
Im lost after that, i dont know how to factor that.
(the "10's" should be lowered)