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I am having trouble sorting out where to begin with solving for unknown value in this equation:

$16^{5a−1} \times 256^{3a} = 128$.

I imagine I would need to change to logarithmic form, but am perplexed by the lack of same base, because if I rearrange into log form:

$\log_{16}(128) \times \log_{16}(128) = (5a-1)+3a$.

Is this an incorrect start? I am thinking that there is a problem with having 128 as N value when it is the product of multiplying two terms. Any directions are much appreciate!

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  • $\begingroup$ You can change any exponential form to any base by doing $a^b = x^{b\log_x(a)}$...in this case it's easier: $256 = 16^2$...then it's just a matter of writing $\left(16^2\right)^{3a} = 16^{\text{?}}$. In this case, it might be even easier to use base $2$ since $16 = 2^4$, $256 = 2^8$, and $128 = 2^7$. $\endgroup$ – Jared May 17 '16 at 2:13
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The use of logs is actually unnecessary here, since by coincidence all bases in the equation can be rewritten as a power of a common base (namely, $2$). We can use exponent laws to simply both sides into the form $2^{X} = 2^Y$ and infer that $X = Y$. Indeed, observe that: \begin{align*} 16^{5a - 1} \cdot 256^{3a} &= 128 \\ (2^4)^{5a - 1} \cdot (2^8)^{3a} &= (2^7) \\ 2^{20a - 4} \cdot 2^{24a} &= 2^7 \\ 2^{44a - 4} &= 2^7 \\ 44a - 4 &= 7 \end{align*} and so on.

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  • $\begingroup$ Thanks Adriano, I started thinking along this line last night and totally missed that 2^7 = 128 ... this definitely helped. Can you recommend any secondary sources that touch on additional laws of exponents outside of our basic multiplication and division ones etc ? $\endgroup$ – Mark Samuel May 19 '16 at 18:59
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$16^{5a−1} \times 256^{3a} =16^{5a-1}\times 16^{2(3a)}=16^{5a+6a-1}=16^{11a-1}=128$.

Now, take a logarithm with base $16$.

This yields $11a-1=\dfrac{7}{4}$. Now, you can solve this for $a$.

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