# How would I solve this exponential equation?

2^x = 7^(x^2)

the (x^2) in the 7 is throwing me off

thanks!

### 7 Answers

- Jeff AaronLv 73 weeks ago
x = (ln(2) +/- sqrt(ln^2(2) + 8*i*pi*n*ln(7))) / (2*ln(7)), for any integer n

- ?Lv 73 weeks ago
2^x = 7^(x^2)

log₂ 2^x = log₂ 7^(x^2)

x log₂ 2 = (x^2) log₂ 7

x (1) = (x^2) log₂ 7

x

---- = log₂ 7

x^2

1

-- = log₂ 7

x

x log₂ 7 = 1

...........1

x = -----------

.......log₂ 7

.....// Change log base

...................log 7

.....log₂ 7 = -------

...................log 2

x = 1 / (log 7 / log 2)

x = log 2 / log 7

x = 0.356207...............ANS

- az_lenderLv 73 weeks ago
x*log(2) = (x^2)*log(7).

One solution is x = 0.

The other solution is the solution to

1*log(2) = x*log(7) =>

x = log(2)/log(7) = around 0.3562.

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- 3 weeks ago
2^x = 7^(x^2)

x * ln(2) = (x^2) * ln(7)

0 = ln(7) * x^2 - ln(2) * x

0 = x * (ln(7) * x - ln(2))

x = 0

ln(7) * x - ln(2) = 0

ln(7) * x = ln(2)

x = ln(2) / ln(7)

x = 0 , ln(2)/ln(7)

There you go.

- hfshawLv 73 weeks ago
You have:

2^x = 7^(x²)

Take the logarithm of both sides:

x*ln(2)= (x²)*ln(7)

By inspection, one solution is x = 0. Having accounted for this solution, we can divide both sides by x:

ln(2) = x*ln(7)

x = ln(2)/ln(7)

So the solutions are x = 0, ln(2)/ln(7).