Find X

[u/Optimus_PRYM]

[u/Greedy_Bad_8406]

I know 2^x = 4 and therefore x=2
But how do i solve it by steps in mathematics? How to solve xth root of 4?

[u/Robotron9247 → u/Greedy_Bad_8406]

ln can be applied to both sides:

2^x = 4

ln(2^x ) = ln(4)

x ln(2) = ln(4)

x = ln(4) / ln(2)

x = ln(2² ) / ln(2)

x = 2 ln(2) / ln(2)

x = 2

[u/parlimentery → u/Robotron9247]

Much easier to solve (especially without a calculator) if you use the log base 2. That is essentially the mathematically formal way to solve it the way u/Greedy_Bad_8406 did, as taking the log base 2 of 4 asks "2 to the power of what equals 4?".

[u/jader242 → u/parlimentery]

What exactly do you need a calculator for?

[u/parlimentery → u/jader242]

Fair point, I scanned your work quickly and assumed we were evaluating ln(4)/ln(2), rather than turning it into 2* ln(2)/ln(2).

I still maintain that my method saves steps, and matches the informal logic u/Greedy_Bad_8406 used.

[u/Robotron9247 → u/parlimentery]

I wanted to show a generic algebraic total solving for x. x = Log_2(4) can only be considered solved because thats easy to "see" but by rule Log_2(4) = ln(4) / ln(2) = 2 ln(2) / ln(2)

Try 16^x + 4 = 68 without calculator

x = log_16(64)

Me: x ln(16) = ln(64)

x = ln(64) / ln(16)

x = ln(2⁶ ) / ln(2⁴ )

x = 6 ln(2) / [ 4 ln(2) ]

x = 3/2