The value of x is -1.596
Solution:
Given equation is:
[tex]4^{(x+3)} = 7[/tex]
Let us solve using change of base formula log base b of y equals log y over log b
From given,
[tex]4^{(x+3)} = 7[/tex]
[tex]\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)[/tex]
Therefore,
[tex]\ln \left(4^{x+3}\right)=\ln \left(7\right)[/tex]
[tex]\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)[/tex]
[tex]\ln \left(4^{x+3}\right)=\left(x+3\right)\ln \left(4\right)\\\\\left(x+3\right)\ln \left(4\right)=\ln \left(7\right)\\[/tex]
Let us simplify the above
[tex]\left(x+3\right)\cdot \:2\ln \left(2\right)=\ln \left(7\right)\\\\\mathrm{Divide\:both\:sides\:by\:}2\ln \left(2\right)\\\\\frac{\left(x+3\right)\cdot \:2\ln \left(2\right)}{2\ln \left(2\right)}=\frac{\ln \left(7\right)}{2\ln \left(2\right)}\\\\[/tex]
[tex]\mathrm{Simplify}\\\\x+3=\frac{\ln \left(7\right)}{2\ln \left(2\right)}\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x+3-3=\frac{\ln \left(7\right)}{2\ln \left(2\right)}-3\\\\\mathrm{Simplify}\\\\x=\frac{\ln \left(7\right)}{2\ln \left(2\right)}-3[/tex]
Substitute the values
ln 7 = 1.9459101490553132
ln 2 = 0.6931471805599453
Therefore,
[tex]x = \frac{1.9459101490553132}{2 \times 0.6931471805599453} - 3\\\\x = 1.40367746103 - 3\\\\x = -1.59632253897 \approx -1.596[/tex]
Thus solution for x is found
