# Thread: Logarithms and Exponents Practice Test - Is my answer right?

1. ## Logarithms and Exponents Practice Test - Is my answer right?

1. So there is a problem question (I'm just going to make things shorter and quicker) and the question is apparently log4 (64) + log1/5 (9) =

My answer is this, I turned 1/5 to 0.2 (I find it easier to solve whole numbers than fractions), then I go log(64) / log(4) = 3? Is this right? (4x4x4=64)
then I go log(9) / log(0.2) = -1.365? Is this right? So I guess 3 + (-1.365) = 1.635 is the answer to this problem? I feel like I miss something... I think...

2. Here is another one 6x = (1/216)^3-x, my answer is x = -1

Edit: Thanks Skeeter, sorry about question number 2, I forgot to put the 3-x, anyway I got the answer, its 4.5

3. And then there is this weird question that I can't seem to figure out the steps, it says

"The growth of a plant during an eight-month period is modelled by the exponential function f(x) = 3(1.25)x Where f(x)represents the height, in centimeters, at the end of each month and x represents the time in months. Determine when the plant will reach a height of 12 cm, to the nearest tenth of a month."

How can I figure this out when its in letters? Not even my calculator can understand it? What sorcery is this?!

2. ## Re: Logarithms and Exponents Practice Test - Is my answer right?

1. 1.635 is a good calculation

2. $6^{-3}=\dfrac{1}{216}$

3. $3(1.25)^x = 12$

$1.25^x = 4$

$x \log(1.25) = \log(4)$

finish it ...

3. ## Re: Logarithms and Exponents Practice Test - Is my answer right?

Just to be clearer on #2, no you're not correct.
=(6^-3)^(3-x) = 6^(-3*(3-x)) = 6^(-9+3x)
so now your exponents of -9+3x must also equal x