We can expand the logarithm of a product as a sum of logarithms:
[tex]\log_dabc=\log_da+\log_db+\log_dc[/tex]
Then using the change of base formula, we can derive the relationship
[tex]\log_xy=\dfrac{\ln y}{\ln x}=\dfrac1{\frac{\ln x}{\ln y}}=\dfrac1{\log_yx}[/tex]
This immediately tells us that
[tex]\log_dc=\dfrac1{\log_cd}=\dfrac12[/tex]
Notice that none of [tex]a,b,c,d[/tex] can be equal to 1. This is because
[tex]\log_1x=y\implies1^{\log_1x}=1^y\implies x=1[/tex]
for any choice of [tex]y[/tex]. This means we can safely do the following without worrying about division by 0.
[tex]\log_db=\dfrac{\ln b}{\ln d}=\dfrac{\frac{\ln b}{\ln c}}{\frac{\ln d}{\ln c}}=\dfrac{\log_cb}{\log_cd}=\dfrac1{\log_bc\log_cd}[/tex]
so that
[tex]\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23[/tex]
Similarly,
[tex]\log_da=\dfrac{\ln a}{\ln d}=\dfrac{\frac{\ln a}{\ln b}}{\frac{\ln d}{\ln b}}=\dfrac{\log_ba}{\log_bd}=\dfrac{\log_db}{\log_ab}[/tex]
so that
[tex]\log_da=\dfrac{-\frac23}{\frac89}=-\dfrac34[/tex]
So we end up with
[tex]\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}[/tex]
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Another way to do this:
[tex]\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}[/tex]
[tex]\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}[/tex]
[tex]\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12[/tex]
Then
[tex]abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}[/tex]
So we have
[tex]\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}[/tex]
