1. ## Help with logs please

We just started doing some work on logs and i cant work out this problem, i think its quite easy but im crap at these.

Two curves:

y=a^x and y=2b^x

Prove that x coord of the point of intersection is 1 / (log(of base2)A-log(of base2)B)

Help would be appreciated

2. y=a^x and y=2b^x

y = a^x ----(1)
y = 2(b^x) ----(2)

At their point of intersection, the two logarithmic curves have the same coordinates.
So, y from (1) = y from (2)
a^x = 2(b^x) ----(3)

Since we are looking for the x-coordinate of the intersection, then we solve for x.
Take the logs, (log to the base 10), of both sides of (3),
x*log(a) = log(2) +x*log(b)
Collect the x-terms,
x*log(a) -x*log(b) = log(2)
x[log(a) -log(b)] = log(2)
Divide both sides by [log(a) -log(b)],
x = log(2) / [log(a) -log(b)] -----(4)

(4) is on base 10.
Since you want x in base 2, then we transform the logs in 4 into logs to the base 2.

You still remember how to change bases for logarithms?
log(base a) to log(base b).

log(base a)(N) = [log(base b)(N)] / [log(base b)(a)] ---***

So,
>>>log(base 10)(2)
= [log(base 2)(2)] / [log(base 2)(10)]
= 1 / [log(base 2)(10)]

>>>log(base 10)(a)
= [log(base 2)(a)] / [log(base 2)(10)]

>>>log(base 10)(b)
= [log(base 2)(b)] / [log(base 2)(10)]

Substitute those into (4),

x = log(2) / [log(a) -log(b)] -----(4)

x = log(base 10)(2) / [log(base 10)(a) -log(base 10)(b)]

x = {1 / log(base 2)(10)} / {[log(base 2)(a) / log(base 2)(10)] -[log(base 2)(b) / log(base 2)(10)]}

Combine the denominator into one fraction only,
x = {1 / log(base 2)(10)} / {[log(base 2)(a) -log(base 2)(b)] / log(base 2)(10)}

Reverse the denominator, to multiply it to the numerator,
x = {1 / log(base 2)(10)} * {log(base 2)(10) / [log(base 2)(a) -log(base 2)(b)]}

The log(base 2)(10) cancels out,
x = 1 / [log(base 2)(a) -log(base 2)(b)] ----answer.

===========
Or, from (3), we can take the logs to the base 2 directly----without passing through the logs to the base 10.

a^x = 2(b^x) ----(3)
log(base 2)(a^x) = log(base 2)(2 b^x)
x*log(base 2)(a) = log(base 2)(2) +x*log(base 2)(b)
x*log(base 2)(a) = 1 +x*log(base 2)(b)
x*log(base 2)(a) -x*log(base 2)(b) = 1
x[log(base 2)(a) -log(base 2)(b)] = 1
x = 1 / [log(base 2)(a) -log(base 2)(b)] ----answer.

3. Ah ok!

Thanks very much for your help