#### Answer

$M=M_0\left(10^{\frac{6-m}{2.5}}\right)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
m=6-2.5\log \left( \dfrac{M}{M_0} \right)
,$ in terms of $
M
,$ use the properties of equality to isolate the logarithmic expression. Then change to exponential form. Finally use again the properties of equality to isolate the needed variable.
$\bf{\text{Solution Details:}}$
Using the properties of equality, the equation above is equivalent to
\begin{array}{l}\require{cancel}
2.5\log \left( \dfrac{M}{M_0} \right)=6-m
\\\\
\dfrac{2.5\log \left( \dfrac{M}{M_0} \right)}{2.5}=\dfrac{6-m}{2.5}
\\\\
\log \left( \dfrac{M}{M_0} \right)=\dfrac{6-m}{2.5}
.\end{array}
Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to
\begin{array}{l}\require{cancel}
\log_{10} \left( \dfrac{M}{M_0} \right)=\dfrac{6-m}{2.5}
\\\\
\dfrac{M}{M_0}=10^{\frac{6-m}{2.5}}
.\end{array}
Using the properties of equality to isolate the needed variable results to
\begin{array}{l}\require{cancel}
M_0\cdot\dfrac{M}{M_0}=M_0\cdot10^{\frac{6-m}{2.5}}
\\\\
M=M_0\left(10^{\frac{6-m}{2.5}}\right)
.\end{array}