QUESTION IMAGE
Question
solving exponential and logarithmic equations
identificacion de soluciones aproximadas
encuentra las soluciones a la ecuación $10^{2x}+11=(x + 6)^2-2$ ¿qué valores son soluciones aproximadas de la ecuación? elige dos respuestas correctas.
-7.4
-4.6
-9.6
-2.4
0.6
To solve the equation \(10^{2x}+11=(x + 6)^2-2\), we can test each given value by substituting it into the equation and checking if both sides are approximately equal.
Step 1: Test \(x=-7.4\)
- Left - hand side (LHS): \(10^{2\times(-7.4)}+11=10^{- 14.8}+11\approx0 + 11=11\)
- Right - hand side (RHS): \((-7.4 + 6)^2-2=(-1.4)^2-2=1.96 - 2=-0.04\)
- Since \(11
eq - 0.04\), \(x = - 7.4\) is not a solution.
Step 2: Test \(x=-4.6\)
- LHS: \(10^{2\times(-4.6)}+11=10^{-9.2}+11\approx0 + 11 = 11\)
- RHS: \((-4.6+6)^2-2=(1.4)^2-2=1.96 - 2=-0.04\)
- Since \(11
eq - 0.04\), \(x=-4.6\) is not a solution.
Step 3: Test \(x=-9.6\)
- LHS: \(10^{2\times(-9.6)}+11=10^{-19.2}+11\approx0 + 11=11\)
- RHS: \((-9.6 + 6)^2-2=(-3.6)^2-2=12.96-2 = 10.96\approx11\)
- So, \(x=-9.6\) is a solution.
Step 4: Test \(x=-2.4\)
- LHS: \(10^{2\times(-2.4)}+11=10^{-4.8}+11\approx0+11 = 11\)
- RHS: \((-2.4 + 6)^2-2=(3.6)^2-2=12.96 - 2=10.96\approx11\)
- So, \(x = - 2.4\) is a solution.
Step 5: Test \(x = 0.6\)
- LHS: \(10^{2\times0.6}+11=10^{1.2}+11\approx15.8489+11=26.8489\)
- RHS: \((0.6 + 6)^2-2=(6.6)^2-2=43.56-2 = 41.56\)
- Since \(26.8489
eq41.56\), \(x = 0.6\) is not a solution.
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-9.6, -2.4