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Due to limitations with browsers mathematical operators (especially + =) get stripped from questions (leaving questions with not enough information to answer them) and it is not entirely clear what the log4 bit means.

I guess that the log4 bit is logarithms to base 4 of 2x^16 (which I'll write as log_4(2x^16) for brevity). If this is so, use normal algebraic operations to make log_4(2x^16) the subject of the equation.

With logs there are useful rules; given 2 numbers 'a' and 'b':

log(ab) = log(a) + log(b)

log(a^b) = b × log(a)

Which means:

log_4(2x^16) = log_4(2) + log_4(x^16) = log_4(2) + 16 × log(x)

and the equation can be further rearranged:

log_4(2x^16) = <whatever>

→ log_4(2) + 16 × log(x) = <whatever>

→ log(x) = (<whatever> - log_4(2)) / 16

Logarithms tell you the power to which the base of the logarithm must be raised to get its argument, for example when using common logs: lg 100 = 2 since 10 must be raised to the power 2 to get 100, ie 10² = 100. (lg is the abbreviation for logs to base 10; ln, or natural logs, is the abbreviation for logs to the base e.) With logs to base 4, it is 4 that is raised to the power of the log to get the original value. eg log_4(16) = 2 since 4^2 = 16.

log_4(2) can be worked out:

The log to any base of the base is 1 (since any number to the power 1 is itself).

Now 2 × 2 = 2² = 4.

→ log_4(4) = 1

→ log_4(2²) = 1

→ 2 × log_4(2) = 1

→ log_4(2) = ½

→ log(x) = (<whatever> - ½) / 16

Back to the rearranged equation; with logs to base 4, if you make both sides the power of 4 you'll get:

4^(log_4(x)) = 4^(<whatever>)

→ x = 4^(<whatever>)

which now solves for x.

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โˆ™ 2015-07-31 05:51:33
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Q: How do you solve the equation 5.2 log4 2x16?
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