How to Add Logarithms with different bases?

Log4 32 + Log2 32

ok jacob s thanks but how? i dont know how you got any of that

4 Answers

  • In general, you need convert one of the logarithms to the same base as the other one by using the relationship:

    log_a(b) * log_b(x) = log_a(x)

    or, written another way:

    log_b(x) = log_a(x)/log_a(b)

    where log_a and log_b mean the logarithms in base a and b, respectively.

    For instance:

    log_4(32) = log_2(32)/log_(4)

    but log_2(4) is 2 (i.e., 2^2 = 4), so:

    log_4(32) = (1/2)*log_2(32)

    You can now use this to carry out the addition:

    log_4(32) + log_2(32) = (1/2)log_2(32) + log_2(32)

    = 1.5*log_2(32)

    = 1.5*log_2(2^5) = 1.5*5 = 7.5

    In this case, however, there is an easier way. Note that 32 = 2^5, and 4 = 2^2, so 32 = 4^2.5.

    The log_2(2^5) = 5, and log_4(4^2.5) = 2.5, so

    log_4(32) + log_2(32) = 5 + 2.5 = 7.5

  • Addition Of Logarithms

  • Use this formulation to calculate logarithm to any base log_b x=(log_k x )/(log_k b) So log_3 80 one=(log_10 80 one )/(log_10 3 )= 4 (you are able to calculate log to base 10 via using your calculator now) and in addition log_4 sixty 4=(log_10 sixty 4 )/(log_10 4 )=3 4+3=7(answer)

  • log32/log4 +log32/log2=

    5/2 + 5= 15/2

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