Respuesta :
We'll use method 1: Write exponents using the same base.
Given equation:
[ -3(10)^{4-x} - 4 = -91 ]First, let's isolate the exponential term: [ -3(10)^{4-x} = -91 + 4 ][ -3(10)^{4-x} = -87 ]Next, let's divide both sides by -3: [ (10)^{4-x} = \frac{-87}{-3} ][ (10)^{4-x} = 29 ]Now, we'll rewrite 29 as a power of 10: [ 10^{4-x} = 10^{\log_{10}29} ][ 10^{4-x} = 10^{\log_{10}29} ]Now, since the bases are the same, the exponents must be equal: [ 4-x = \log_{10}29 ]Now, we solve for x: [ x = 4 - \log_{10}29 ]So, ( x = 4 - \log_{10}29 ).
Answer:
x â 2.5376
Step-by-step explanation:
To solve the equation [tex]\sf -3 \times 10^{4-x} - 4 = -91[/tex], we'll utilize the following logarithmic rule/property:
[tex]\sf a \cdot b^x = a \cdot e^{x \cdot \log(b)}[/tex]
Using this rule, we can rewrite the equation as:
[tex]\sf -3 \cdot e^{(\log(10) \cdot (4-x))} - 4 = -91[/tex]
Now, notice that [tex]\sf \ln(10) = 1[/tex] because [tex]\sf \ln(10)[/tex] is the natural logarithm of [tex]\sf e[/tex] raised to the power of [tex]\sf 1[/tex], which is simply [tex]\sf 10[/tex].
So, the equation simplifies to:
[tex]\sf -3 \cdot e^{(4-x)} - 4 = -91[/tex]
Next, we'll add [tex]\sf 4[/tex] to both sides to isolate the exponential term:
[tex]\sf -3 \cdot e^{(4-x)} = -91 + 4[/tex]
[tex]\sf -3 \cdot e^{(4-x)} = -87[/tex]
Now, divide both sides by [tex]\sf -3[/tex] to isolate the exponential term:
[tex]\sf e^{(4-x)} = \dfrac{-87}{-3}[/tex]
[tex]\sf e^{(4-x)} = 29[/tex]
Now, we take the natural logarithm of both sides to eliminate the exponential term:
[tex]\sf \log(e^{(4-x)}) = \log(29)[/tex]
[tex]\sf 4 - x = \log(29)[/tex]
Finally, solve for [tex]\sf x[/tex]:
[tex]\sf 4 - x = \log(29)[/tex]
[tex]\sf -x = \log(29) - 4[/tex]
[tex]\sf x = 4 - \log(29)[/tex]
[tex]\sf x \approx 4 - 1.4623979978989 [/tex]
[tex]\sf x \approx 2.5376020021010 [/tex]
[tex]\sf x \approx 2.5376 \textsf{(in 4 d.p.)}[/tex]
Therefore, x â 2.5376.