Math problems

[I'm clever]

I'm stuck on these three maths questions.

1) In a geometric progression, the sum to infinity is four times the first term.

(i) Show that the common ratio is 3
(ii) Given that the third term is 9, find the first term.
(iii) Find the sum of the first twenty terms.


2) Solve the equation log_2 x + 2 log_2 3 = log_2(x + 5).

3) Find:

200
Σ (3n+2)
n=101

for 3) Should I subtract the series cause it doesn't start with 1?

[eumyang]

Show us what you have tried. (We can't just give answers here.) For #2 you'll need to use the logarithmic property
\log_b xy = \log_b x + \log_b y
... among others.

[I'm clever]

Show us what you have tried. (We can't just give answers here.) For #2 you'll need to use the logarithmic property
\log_b xy = \log_b x + \log_b y
... among others.

\log_2 x + 2\log_2 3 = \log_2 (x+5)

Power rule:

\log_2 x + \log_2 3^2 = \log_2 (x+5)

Addition rule:

\log_2 (x \times 3x^2) = \log_2 (x+5)

\log_2 3x^2 = \log_2 (x+5)

[I'm clever]

\log_2 x + 2\log_2 3 = \log_2 (x+5)

Power rule:

\log_2 x + \log_2 3^2 = \log_2 (x+5)

Addition rule:

\log_2 (x \times 3x^2) = \log_2 (x+5)

\log_2 3x^2 = \log_2 (x+5)

Is this right?

[eumyang]

\log_2 x + 2\log_2 3 = \log_2 (x+5)

Power rule:

\log_2 x + \log_2 3^2 = \log_2 (x+5)

Addition rule:

\log_2 (x \times 3x^2) = \log_2 (x+5)

Where did that second "x" come from? It should be
\log_2 (x \times 3^2) = \log_2 (x+5)
or
\log_2 (9x) = \log_2 (x+5)

After this, use the property:
if logb x = logb y, then x = y
... and solve for x.

[eumyang]

I'm stuck on these three maths questions.

1) In a geometric progression, the sum to infinity is four times the first term.

(i) Show that the common ratio is 3
Are you sure you copied the problem right? My understanding is that unless r < 1 the series won't converge. If the sum to infinity is four times the first term, then I'm getting 3/4 as the common ratio, not 3.