Answer:
x ≈ 0.83365
Step-by-step explanation:
To solve the equation [tex]-7 \times 10^{8-10x} + 9 = 4[/tex], we'll use the following logarithmic rule:
[tex]a \cdot b^x = a \cdot e^{x \cdot \ln(b)}[/tex]
Using this rule, we can rewrite the equation as:
[tex]-7 \cdot e^{\ln(10) \cdot (8-10x)} + 9 = 4[/tex]
Now, observe that [tex]\ln(10) = 1[/tex] because [tex]\ln(10)[/tex] is the natural logarithm of [tex]e[/tex] raised to the power of [tex]1[/tex], which is just [tex]10[/tex]. So, the equation simplifies to:
[tex]-7 \cdot e^{8-10x} + 9 = 4[/tex]
Now, we subtract 9 from both sides to isolate the exponential term:
[tex]-7 \cdot e^{8-10x} = 4 - 9[/tex]
[tex]-7 \cdot e^{8-10x} = -5[/tex]
Next, we divide both sides by [tex]-7[/tex] to isolate the exponential term:
[tex]e^{8-10x} = \dfrac{-5}{-7}[/tex]
[tex]e^{8-10x} = \dfrac{5}{7}[/tex]
Now, we take the natural logarithm of both sides to eliminate the exponential term:
[tex]\ln(e^{8-10x}) = \ln\left(\dfrac{5}{7}\right)[/tex]
[tex]8 - 10x = \ln\left(\dfrac{5}{7}\right)[/tex]
Finally, we solve for [tex]x[/tex]:
[tex]8 - 10x = \ln\left(\dfrac{5}{7}\right)[/tex]
[tex]-10x = \ln\left(\dfrac{5}{7}\right) - 8[/tex]
[tex]x = \dfrac{\ln\left(\dfrac{5}{7}\right) - 8}{-10}[/tex]
[tex] x \approx 0.8336472236621 [/tex]
[tex] x \approx 0.83365 \textsf{(in 5 d.p.)}[/tex]
Therefore, x ≈ 0.83365.
