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\require{AMSmath}

Uitwerkingen van de opgaven

Opgave 1

$
\eqalign{
& 4^{x - 2} \cdot 8^{3x - 1} = \sqrt 2   \cr
& \left( {2^2 } \right)^{x - 2} \cdot \left( {2^3 } \right)^{3x - 1} = 2^{\frac{1}
{2}}   \cr
& 2^{2x - 4} \cdot 2^{9x - 3} = 2^{\frac{1}
{2}}   \cr
& 2^{2x - 4 + 9x - 3} = 2^{\frac{1}
{2}}   \cr
& 2^{11x - 7} = 2^{\frac{1}
{2}}   \cr
& 11x - 7 = \frac{1}
{2} \cr
& 11x = 7\frac{1}
{2} \cr
& x = \frac{{15}}
{{22}} \cr}
$

$
\eqalign{
& 9 \cdot 3^{2x} - 3^{x + 2} = 54 \cr
& 9 \cdot 3^{2x} - 9 \cdot 3^x = 54 \cr
& 3^{2x} - 3^x = 6 \cr
& \left( {3^x } \right)^2 - 3^x - 6 = 0 \cr
& y^2 - y - 6 = 0 \cr
& (y - 3)(y + 2) = 0 \cr
& y = 3\,\,of\,\,y = - 2 \cr
& 3^x = 3\,\,of\,\,3^x = - 2\,\,(kan\,\,niet) \cr
& x = 1 \cr}
$

$
\eqalign{
& 3^{x^2 } = \left( {\frac{1}
{3}} \right)^{ - x - 2}   \cr
& 3^{x^2 } = \left( {3^{ - 1} } \right)^{ - x - 2}   \cr
& 3^{x^2 } = 3^{x + 2}   \cr
& x^2 = x + 2 \cr
& x^2 - x - 2 = 0 \cr
& (x - 2)(x + 1) = 0 \cr
& x = 2 \vee x = - 1 \cr}
$

$
\eqalign{
& \left( {\frac{1}
{2}\sqrt 2 } \right)^{2x + 3} = 2^x   \cr
& \left( {2^{ - 1} \cdot 2^{\frac{1}
{2}} } \right)^{2x + 3} = 2^x   \cr
& \left( {2^{ - \frac{1}
{2}} } \right)^{2x + 3} = 2^x   \cr
& 2^{ - x - 1\frac{1}
{2}} = 2^x   \cr
& - x - 1\frac{1}
{2} = x \cr
& 2x = - 1\frac{1}
{2} \cr
& x = - \frac{3}
{4} \cr}
$

Opgave 2

a.

$
\eqalign{
& 2^{x - 3} \cdot 5^{x + 1} = 62,5 \cr
& 2^{ - 4} \cdot 2^{x + 1} \cdot 5^{x + 1} = 62,5 \cr
& 2^{x + 1} \cdot 5^{x + 1} = 1000 \cr
& 10^{x + 1} = 1000 \cr
& 10^{x + 1} = 10^3   \cr
& x + 1 = 3 \cr
& x = 2 \cr}
$

b.

$
\eqalign{
& 16^x = 8 + 7 \cdot 4^x   \cr
& 4^{2x} = 8 + 7 \cdot 4^x   \cr
& {\text{Neem}}\,\,\,t = 4^x   \cr
& t^2 = 8 + 7t \cr
& t^2 - 7t - 8 = 0 \cr
& (t + 1)(t - 8) = 0 \cr
& t = - 1\,\,{\text{of}}\,\,t = 8 \cr
& 4^x = - 1\,\,\,{\text{(kan}}\,\,{\text{niet)}}\,\,{\text{of}}\,\,4^x = 8 \cr
& 2^{2x} = 2^3   \cr
& 2x = 3 \cr
& x = 1\frac{1}
{2} \cr}
$