Simplify $\log((10\cdot 8 )^{\frac{1}{2}} \times (0.24)^{\frac{5}{3}} \div (90)^{-2})$
$\Rightarrow \log(10\cdot 8)^{\frac{1}{2}}+\log(0.24)^{\frac{5}{3}}-\log(90)^{-2} \tag{1}$
$\Rightarrow \dfrac{1}{2}(\log10+3\log2)+\dfrac{5}{3}(\log(\dfrac{6}{25}))+2\log(2\cdot5\cdot3^2) \tag{2}$
$\Rightarrow \dfrac{1}{2}(1+3\log2)+\dfrac{5}{3}(\log2+\log3-2\log5)+2(\log2+\log5+2\log3) \tag{3}$
$\Rightarrow \dfrac{1}{2}+\dfrac{3}{2}\log2+\dfrac{5}{3}\log2+\dfrac{5}{3}\log3-\dfrac{10}{3}\log5+2\log2+2\log5+4\log3 \tag{4}$
$\Rightarrow \dfrac{1}{2}+\dfrac{31}{6}\log2+\dfrac{17}{3}\log3-\dfrac{4}{3}\log5 \tag{5}$
$\Rightarrow \dfrac{1}{2}+\dfrac{31}{6}\log2+\dfrac{17}{3}\log3-\dfrac{4}{3}(1-\log2) \tag{6}$
$\Rightarrow -\dfrac{5}{6}+\dfrac{39}{6}\log2+\dfrac{17}{3}\log3 \tag{7}$
But the answer is given as $-\dfrac{11}{6}+6\log2+\dfrac{43}{6}\log3$. My calculations look all good to me, but clearly somewhere it is wrong. Thanks for the help.
If it's $10.8$ (i.e. $10 \frac 45$) rather than $10\cdot 8$ (i.e $80$)
You'd get $\log_{10} 10.8^{\frac 12} = \frac 12(\frac {4\cdot 3^3}{10})=\frac 12(2\log 2 + 3\log 3 - 1)= \log 2 + \frac 32 \log 3 - \frac 12$
rather than $\log 80^{\frac 12} =\frac 12( \log 8\cdot 10) = \frac 32\log 2 + \frac 12$
If so your answer is off by $-\frac 12\log 2 +\frac 32\log 3 - 1$.
Add that to your answer and you get $[-\dfrac{5}{6}+\dfrac{39}{6}\log2+\dfrac{17}{3}\log3] + [ -\frac 12\log 2 +\frac 32\log 3 - 1]=$ $-\frac {11}6 + 6\log 2 +\frac {43}6\log 3$ which is the texts answer.
So that's that, I presume. You interpreted a decimal point as a times sign. (And had a lot more patience and lack of errors than I ever could manage.)