`10^(6x-1)=2^(2x+1)`

Find x.

Expert Answers

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Take the log (base10) of each side.  That gives:

Log(10^6x-1) = Log(2^2x+1)

6x-1 = (2x+1)log 2

6x-1 = 0.301(2x+1)

6x-1 = 0.602x+0.301

Subtract 0.602x from each side:

5.398x - 1 = 0.301

Add 1 to each side

5.398x = 1.301

Divide by 5.398:

x = 0.241

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Use the formulas `x^(a-b)=x^a/x^b` and `x^(a+b)=x^ax^b`

`10^(6x)/10=2^(2x)(2)`

Use the formula `x^(ab)=x`

`10^(6x)=(20)(2^(2x))`

`log(10^(6x))=log((20)(2^(2x)))`

Use the formula `log(ab)=loga+logb`

`6x=log_10(20)+log_10(2^(2x))`       (1)

Substitute 10 for a, 2^(2x) for N and 2 for b in the formula `log_a(N)=(log_b(N))/(log_b(a))`

`log_10(2^(2x))=(log_2(2^(2x)))/(log_2(10))=(2x)/(log_2(10)=`       (2)

Substitute (2) in (1)

`6x=log_10(20)+(2x)/(log_2(10))`

`(6-2/(log_2(10)))x=log_10(20)`

`x=(log_10(20))/(6-2/(log_2(10))`

Evaluate the expression.

`x=1.3/(6-2/3.32)=1.3/(6-0.6)=1.3/5.4=0.24`

Thus x=0.24

Verify graphically: x=0.24 is the point where `y=10^(6x-1)` and `y=2^(2x+1)` intersect.

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