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Thread: Logarithm

  1. #1
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    Logarithm

    solve the equation
    (2^x)(3^(x+1)=10
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  2. #2
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    Quote Originally Posted by Rose Wanjohi View Post
    solve the equation
    (2^x)(3^(x+1)=10
    Assuming the question is $\displaystyle (2^x)(3^{x+1}) = 10$

    Hint: $\displaystyle 3^{x+1} = 3 \times 3^x$

    I get an answer of 0.672 (3sf).

    I've put my working in a spoiler below so you can check your own too

    Spoiler:
    $\displaystyle (3)(2^x)(3^x) = 10$

    Take the logs (base 10 will keep it simpler) $\displaystyle log(3 \times 2^x \times 3^x) = 1$

    Use the law $\displaystyle log_b(ac) = log_b(a) + log_b(c)$ to separate the left hand side

    $\displaystyle log(3) + log(2^x) + log(3^x) = 1$

    Take log(3) from both sides and then use the law which says $\displaystyle log(a^k) = k\, log(a)$

    $\displaystyle x\, log(2) + x\, log(3) = 1-log(3)$

    Factor out x: $\displaystyle x(log(2)+log(3)) = 1-log(3)$

    Divide through to isolate x: $\displaystyle \frac{1-log(3)}{log(2)+log(3)}$
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  3. #3
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    logarithm

    yeah i got it,
    xlog2+xlog3+log3=1
    log3=0.4771
    log2=0.3010
    therefore 0.3010x+0.4771x+0.4771=1
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