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.

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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.