Question

1.Given log - 2 = 0.3010 and log.. 3-0.4771. If 5^x-1= 6^2-x, then the value of 'x' 1) 1.527 2) 1.528 3) 15.28 4) 152.8​

Answer 1

Answer:

The question is "Given log 2=0.3010 and log 3=0.4771, if 5^x-1=6^2-x,then find value of X".

sol: 5^x-1 = 6^2-x

log 5^x-1 = log 6^2-x

(x-1) log10/2=(2-x)(log(3*2))

(x-1)(log10-log2)=(2-x)(0.3010+0.4771)

(x-1)(0.6990)=(2-x)(0.7781)

o.990x-0.6990=1.5562-0.7781x

0.6990x+0.7781x=1.5562+0.6990

1.4771x=2.2552

x = 2.2552/1.4771

x=1.52689 (approx)

OR

5^x-1 = 6^2-x

log 5^x-1=log 6^2-x

(x-1) log10/2 =(2-x) log 3*2

(x-1) (log10-log2) = (2-x) (log2+ log 3)

(x-1) (1-0.3010) = (2-x) (0.3010 + 0.4771)

(x-1) (0.6990) = (2-x) (0.7781)

0.6990x-0.6990=1.5562-0.6990

1.4771x=2.2552

x=2.2552/1.4771

x=1.5268 (approx)

NOTE:

(^ is caret or circumflex and it means to the power of)

Step-by-step explanation: