what is the approximate value of x in the equation below?\nlog₅15 = x + 3\n-2.523\n-1.317\n2.880\n7.485

Answer:

A. -2.523

Explanation:

Step1: Isolate (x)

Given (\log_{5}15=x + 3), we can rewrite it as (x=\log_{5}15-3).

Step2: Use the change - of - base formula

The change - of - base formula is (\log_{a}b=\frac{\ln b}{\ln a}). So (\log_{5}15=\frac{\ln15}{\ln5}). We know that (\ln15=\ln(3\times5)=\ln3+\ln5\approx1.0986 + 1.6094=2.708) and (\ln5\approx1.6094). Then (\frac{\ln15}{\ln5}=\frac{2.708}{1.6094}\approx1.682).

Step3: Calculate (x)

Substitute the value of (\log_{5}15) into (x=\log_{5}15 - 3). So (x\approx1.682-3=-1.318) (using a calculator, (\log_{5}15=\frac{\log15}{\log5}=\frac{1.1761}{0.6990}\approx1.682), (x = 1.682-3=-1.318)).

Wait, there is a mistake above. Let's use the correct formula (x=\log_{5}15-3). Using the change - of - base formula (\log_{5}15=\frac{\ln15}{\ln5}\approx\frac{2.70805}{1.60944}\approx1.6826). Then (x = 1.6826-3=-1.3174\approx - 1.317).

Another way:

Step1: Isolate (x)

From (\log_{5}15=x + 3), we get (x=\log_{5}15-3).

Step2: Use the property (\log_{a}(M)=\frac{\log M}{\log a})

(x=\frac{\log15}{\log5}-3). Since (\log15=\log(3\times5)=\log3+\log5\approx0.4771+0.69897 = 1.17607), (\log5\approx0.69897). Then (\frac{\log15}{\log5}=\frac{1.17607}{0.69897}\approx1.6826).

Step3: Calculate (x)

(x=1.6826 - 3=-1.3174\approx - 1.317)