Thread: Is this equation right? (Logs/Algebra)

1. Is this equation right? (Logs/Algebra)

So the first question is log2 (16) + log2 (8) and what I did at log2 (16) to solve it is 2x2x2x2 = 16, which means the answer is 4. And at log2 8 I did 2x2x2 = 8, which is 3. So 4+3= 7 is my answer, is this right?

Second answer is kinda weird... The equation is log1/2 (1/8) - log1/2 (32), and I don't know how to solve this one because it's in fractions not whole numbers, can someone help me show the steps how? Btw I only have the basic kindergarten calculators. (I know 1/2 is equal to 0.5, but I'm not sure if that would work?)

EDIT: Nevermind! I got the answer! Basically I just changed 1/2 to 0.5 and 1/8 to 0.125, I think I might solve this one!

Third, I'm also having trouble putting this in algebra or logarithmic terms, 8^x-3 = 16^x+3

2. Re: Is this equation right? (Logs/Algebra)

yes
$\log_2(16)+\log_2(8) = 4+3=7$

well done

the second one is $\log_{1/2}(1/8) - \log_{1/2}(32)$ ?

if so note that

$\dfrac 1 8 = \left(\dfrac 1 2\right)^3$

and

$32 = 2^5 = \left(\dfrac 1 2\right)^{-5}$

I assume you mean

$8^{x-3} = 16^{x+3}$

might as well take everything down to powers of 2

$8^{x-3} = 2^{3(x-3)}$

$16^{x+3} = 2^{4(x+3)}$

and now we can say

$3(x-3) = 4(x+3)$

which is easily solved. (by you, or apparently by SlipEternal)

3. Re: Is this equation right? (Logs/Algebra)

Originally Posted by PeanutBrain
So the first question is log2 (16) + log2 (8) and what I did at log2 (16) to solve it is 2x2x2x2 = 16, which means the answer is 4. And at log2 8 I did 2x2x2 = 8, which is 3. So 4+3= 7 is my answer, is this right?

Second answer is kinda weird... The equation is log1/2 (1/8) - log1/2 (32), and I don't know how to solve this one because it's in fractions not whole numbers, can someone help me show the steps how? Btw I only have the basic kindergarten calculators. (I know 1/2 is equal to 0.5, but I'm not sure if that would work?)

EDIT: Nevermind! I got the answer! Basically I just changed 1/2 to 0.5 and 1/8 to 0.125, I think I might solve this one!

Third, I'm also having trouble putting this in algebra or logarithmic terms, 8^x-3 = 16^x+3
For the last one, you have:

$2^{3x}-3=2^{4x}+3$
$2^{3x}=2^{4x}+6$
$0=2^{4x}-2^{3x}+6$
Let $y=2^x$. Then you have
$0=y^4-y^3+6$

That is a 4th degree polynomial with no real solutions. So, your equation has no real solutions.

$8^{x-3}=16^{x+3}$
$2^{3x-9}=2^{4x+12}$
$3x-9=4x+12$
$x=-21$